DIFFERENTIATION MCQs for JEE MATHEMATICS — Medium, Page 5

121 Medium

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Application +4 | -1

If

\tan ^{-1} \frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: B

\frac{1}{2 \sqrt{1-x^2}}

Option B (

\frac{1}{2 \sqrt{1-x^2}}

) is correct because after substituting

x = \cos 2\theta

and simplifying the expression, the derivative is found to be

\frac{1}{2 \sqrt{1-x^2}}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2 \sqrt{1-x^2}}$ . The derivative is found by using a trigonometric substitution and simplifying.

Step-by-Step Solution:

Step 1: Substitute $x = \cos 2\theta$ .
$y = \tan^{-1}(\frac{\sqrt{1 + \cos 2\theta} - \sqrt{1 - \cos 2\theta}}{\sqrt{1 + \cos 2\theta} + \sqrt{1 - \cos 2\theta}})$

Step 2: Use trigonometric identities to simplify the expression.
$1 + \cos 2\theta = 2 \cos^2 \theta$ and $1 - \cos 2\theta = 2 \sin^2 \theta$
$y = \tan^{-1}(\frac{\sqrt{2 \cos^2 \theta} - \sqrt{2 \sin^2 \theta}}{\sqrt{2 \cos^2 \theta} + \sqrt{2 \sin^2 \theta}}) = \tan^{-1}(\frac{\sqrt{2} \cos \theta - \sqrt{2} \sin \theta}}{\sqrt{2} \cos \theta + \sqrt{2} \sin \theta}) = \tan^{-1}(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta})$

Step 3: Divide both the numerator and the denominator by $\cos \theta$ .
$y = \tan^{-1}(\frac{1 - \tan \theta}{1 + \tan \theta})$

Step 4: Recognize the expression as the tangent of a difference.
$y = \tan^{-1}(\tan(\frac{\pi}{4} - \theta)) = \frac{\pi}{4} - \theta$

Step 5: Substitute back $\theta = \frac{1}{2} \cos^{-1} x$ .
$y = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$

Step 6: Differentiate with respect to $x$ .
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{1}{2} \cos^{-1} x) = 0 - \frac{1}{2} (-\frac{1}{\sqrt{1 - x^2}}) = \frac{1}{2 \sqrt{1 - x^2}}$

Why other options are wrong:
- **Option A ( $\frac{1}{\sqrt{1-x^2}}$ ):** This is incorrect because the factor of $\frac{1}{2}$ is missing. This could be due to an error in differentiating or simplifying the expression.
- **Option C ( $\frac{x}{\sqrt{1-x^2}}$ ): This is incorrect because the derivative is not in this form. This could be due to an error in differentiating or simplifying the expression.
- Option D (none of these): This is incorrect because option B is the correct answer.

Key Concepts:**
- Trigonometric substitution
- Trigonometric identities
- Inverse trigonometric functions and their derivatives
- $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

Memory Aids:
- Remember the trigonometric identities for simplifying expressions.
- Recall the derivative of $\cos^{-1}(x)$ .

Quick Tips:
- Use trigonometric substitutions to simplify expressions involving square roots of the form $\sqrt{1 \pm x}$ or $\sqrt{1 \pm x^2}$ .
- Simplify the expression as much as possible before differentiating.

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Analyzing +4 | -1

If

f(x)=\left|\log _e\right| x||

then

f^{\prime}(x)

equals

View answer and explanation

Correct option: B

\frac{1}{x}

for

|x|>1

and

-\frac{1}{x}

for

|x|<1

Option B is correct because it considers the absolute value of both the logarithm and x, leading to different derivative expressions for |x| > 1 and |x| < 1.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{x}}$ for $|x|>1$ and $-\frac{{1}}{{x}}$ for $|x|<1$ . This is because the absolute values affect the sign of the derivative depending on the value of $x$ .

Step-by-Step Solution:

Step 1: Analyze the function $f(x) = |\log_e |x|||$ .

Step 2: Consider the case when $x > 1$ .\nIn this case, $|x| = x$ and $\log_e |x| = \log_e x > 0$ , so $f(x) = \log_e x$ .\nThen, $f'(x) = \frac{{1}}{{x}}$ .

Step 3: Consider the case when $x < -1$ .\nIn this case, $|x| = -x$ and $\log_e |x| = \log_e (-x) > 0$ , so $f(x) = \log_e (-x)$ .\nThen, $f'(x) = \frac{{1}}{{-x}} (-1) = \frac{{1}}{{x}}$ .

Step 4: Consider the case when $0 < x < 1$ .\nIn this case, $|x| = x$ and $\log_e |x| = \log_e x < 0$ , so $f(x) = -\log_e x$ .\nThen, $f'(x) = -\frac{{1}}{{x}}$ .

Step 5: Consider the case when $-1 < x < 0$ .\nIn this case, $|x| = -x$ and $\log_e |x| = \log_e (-x) < 0$ , so $f(x) = -\log_e (-x)$ .\nThen, $f'(x) = -\frac{{1}}{{-x}} (-1) = -\frac{{1}}{{x}}$ .

Step 6: Summarize the results.
$f'(x) = \begin{{cases}} \frac{{1}}{{x}}, & |x| > 1 \\ -\frac{{1}}{{x}}, & |x| < 1 \end{{cases}}$

Why other options are wrong:
- Option A: This is incorrect because it does not account for the absolute value of the logarithm, which changes the sign of the derivative for $|x| < 1$ .
- Option C: This is incorrect because it has the signs reversed for $|x| > 1$ and $|x| < 1$ .
- Option D: This is incorrect because it only considers the sign of $x$ and not the absolute value of the logarithm.

Key Concepts:
- Differentiation of logarithmic functions
- Absolute value functions
- Chain rule

Memory Aids:
- Remember that the derivative of $\log_e x$ is $\frac{{1}}{{x}}$ .
- Consider the cases when $x$ is positive and negative separately when dealing with absolute values.

Quick Tips:
- Always consider the domain of the function when dealing with logarithms and absolute values.
- Break down the function into different cases based on the absolute values.

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Application +4 | -1

If

f(x)=\tan ^{-1}\left(\frac{3 \sin x}{4+5 \cos x}\right)

then

f^{\prime}\left(\frac{\pi}{3}\right)=

View answer and explanation

Correct option: A

\frac{3}{7}

Option A (3/7) is correct because it is the value of the derivative of f(x) evaluated at x = pi/3 after simplification.

Solution approach

Correct Answer + Core Reason: $\frac{3}{7}$ . The derivative of $f(x)$ evaluated at $x = \frac{\pi}{3}$ is $\frac{3}{7}$ .

Step-by-Step Solution:

Step 1: Find the derivative of $f(x) = \tan^{-1}\left(\frac{3 \sin x}{4+5 \cos x}\right)$ .\nUsing the chain rule, $f'(x) = \frac{1}{1 + \left(\frac{3 \sin x}{4+5 \cos x}\right)^2} \cdot \frac{d}{dx} \left(\frac{3 \sin x}{4+5 \cos x}\right)$ .

Step 2: Calculate the derivative of the inner function:
$\frac{d}{dx} \left(\frac{3 \sin x}{4+5 \cos x}\right) = \frac{(4+5 \cos x)(3 \cos x) - (3 \sin x)(-5 \sin x)}{(4+5 \cos x)^2} = \frac{12 \cos x + 15 \cos^2 x + 15 \sin^2 x}{(4+5 \cos x)^2} = \frac{12 \cos x + 15}{(4+5 \cos x)^2}$ .

Step 3: Substitute the derivative back into the expression for $f'(x)$ :
$f'(x) = \frac{1}{1 + \frac{9 \sin^2 x}{(4+5 \cos x)^2}} \cdot \frac{12 \cos x + 15}{(4+5 \cos x)^2} = \frac{(4+5 \cos x)^2}{(4+5 \cos x)^2 + 9 \sin^2 x} \cdot \frac{12 \cos x + 15}{(4+5 \cos x)^2} = \frac{12 \cos x + 15}{16 + 40 \cos x + 25 \cos^2 x + 9 \sin^2 x} = \frac{12 \cos x + 15}{16 + 40 \cos x + 25 \cos^2 x + 9(1 - \cos^2 x)} = \frac{12 \cos x + 15}{25 + 40 \cos x + 16 \cos^2 x}$ .

Step 4: Evaluate $f'(\frac{\pi}{3})$ :
$f'(\frac{\pi}{3}) = \frac{12 \cos(\frac{\pi}{3}) + 15}{25 + 40 \cos(\frac{\pi}{3}) + 16 \cos^2(\frac{\pi}{3})} = \frac{12(\frac{1}{2}) + 15}{25 + 40(\frac{1}{2}) + 16(\frac{1}{4})} = \frac{6 + 15}{25 + 20 + 4} = \frac{21}{49} = \frac{3}{7}$ .

Final Answer: $\frac{3}{7}$

Why other options are wrong:
- **Option B ( $\frac{2}{2 \sqrt{3}+5}$ ): This is incorrect due to an error in the simplification or evaluation of the derivative.
- Option C ( $\frac{6}{5 \sqrt{3}+3}$ ): This is incorrect due to an error in the simplification or evaluation of the derivative.
- Option D ( $\frac{6}{13}$ ): This is incorrect due to an error in the simplification or evaluation of the derivative.

Key Concepts:
- Derivatives of inverse trigonometric functions
- Chain rule
- Trigonometric identities

Memory Aids:**
- Remember the derivative of $\tan^{-1}(x)$ : $\frac{1}{1+x^2}$

Quick Tips:
- Simplify the expression after each step to avoid errors.
- Use trigonometric identities to simplify the expression.

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Analyzing +4 | -1

Let

f \,\theta=\sin \left\{\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right\}

, where

-\frac{\pi}{4} < \theta < \frac{\pi}{4}

. Then the value of

\frac{d}{d(\tan \theta)}(f(\theta))

, is

View answer and explanation

Correct option: A

1

Option A (1) is correct because after simplification, f(θ) = tan θ, and therefore the derivative of f(θ) with respect to tan θ is 1.

Solution approach

Correct Answer + Core Reason: 1. After simplifying the expression, $f(\theta) = \tan \theta$ , so the derivative with respect to $\tan \theta$ is 1.

Step-by-Step Solution:

Step 1: Simplify the expression inside the sine function.
$\tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}}) = \tan^{-1}(\frac{\sin \theta}{\sqrt{1 - 2\sin^2 \theta}})$

Step 2: Rewrite the expression in terms of sine.
$f(\theta) = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\sqrt{1 - 2\sin^2 \theta}} ) \right]$ \nLet $x = \sin \theta$ , then $\sqrt{1 - 2x^2} = \sqrt{x^2 + 1 - 2x^2} = \sqrt{1-x^2 - x^2}$
$f(\theta) = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}}) \right] = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\sqrt{1 - 2\sin^2 \theta}}) \right]$ \nSince $\cos 2\theta = 1 - 2\sin^2 \theta$
$\Rightarrow f(\theta) = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}}) \right] = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\sqrt{1 - 2\sin^2 \theta}}) \right] = \sin \left[ \tan^{-1}(\frac{\sin \theta}{\cos \theta}) \right] = \sin \left[ \tan^{-1}(\tan \theta) \right] = \sin \left[ \theta \right]$ \nThis simplification is incorrect. Let's try another approach.
$\tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}}) = \tan^{-1}(\frac{\sin \theta}{\sqrt{\cos^2 \theta - \sin^2 \theta}})$ \nLet $y = \tan \theta$ , then $\sin \theta = \frac{y}{\sqrt{1+y^2}}$ and $\cos \theta = \frac{1}{\sqrt{1+y^2}}$
$\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \frac{1}{1+y^2} - \frac{y^2}{1+y^2} = \frac{1-y^2}{1+y^2}$
$\sqrt{\cos 2\theta} = \sqrt{\frac{1-y^2}{1+y^2}}$
$\frac{\sin \theta}{\sqrt{\cos 2\theta}} = \frac{\frac{y}{\sqrt{1+y^2}}}{\sqrt{\frac{1-y^2}{1+y^2}}} = \frac{y}{\sqrt{1-y^2}}$
$f(\theta) = \sin \left[ \tan^{-1}(\frac{\tan \theta}{\sqrt{1 - \tan^2 \theta}} ) \right]$ \nThis is also incorrect.
$\tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}}) = \tan^{-1}(\frac{\sin \theta}{\sqrt{1-2\sin^2 \theta}})$ \nLet $x = \tan^{-1}(\frac{\sin \theta}{\sqrt{\cos 2\theta}})$ , then $\tan x = \frac{\sin \theta}{\sqrt{\cos 2\theta}}$
$\sin x = \frac{\tan x}{\sqrt{1+\tan^2 x}} = \frac{\frac{\sin \theta}{\sqrt{\cos 2\theta}}}{\sqrt{1 + \frac{\sin^2 \theta}{\cos 2\theta}}} = \frac{\sin \theta}{\sqrt{\cos 2\theta + \sin^2 \theta}} = \frac{\sin \theta}{\sqrt{\cos^2 \theta - \sin^2 \theta + \sin^2 \theta}} = \frac{\sin \theta}{\cos \theta} = \tan \theta$
$f(\theta) = \sin \left[ \sin^{-1}(\tan \theta) \right] = \tan \theta$

Step 3: Find the derivative of $f(\theta)$ with respect to $\tan \theta$ .
$\frac{d}{d(\tan \theta)}(f(\theta)) = \frac{d}{d(\tan \theta)}(\tan \theta) = 1$

Why other options are wrong:
- Option B: This is incorrect because the derivative is not 2.
- Option C: This is incorrect because the derivative is not 3.
- Option D: This is incorrect because the derivative is not 4.

Key Concepts:
- Inverse trigonometric functions
- Trigonometric identities
- Differentiation

Memory Aids:
- Remember trigonometric identities like $\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 1 - 2\sin^2 \theta$ .

Quick Tips:
- Simplify the expression inside the trigonometric functions before differentiating.

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Application +4 | -1

\frac{d}{{dx}}\left[ {{{\tan }^{ - 1}}\left( {\frac{{a - x}}{{1 + ax}}} \right)} \right]

is equal to

View answer and explanation

Correct option: A

- \frac{1}{{1 + {x^2}}}

Option A (

\frac{{-1}}{{1 + x^2}}

) is correct because the expression inside the derivative simplifies to

\tan^{-1}(a) - \tan^{-1}(x)

, and the derivative of

\tan^{-1}(x)

is

\frac{1}{{1 + x^2}}

.

Solution approach

Correct Answer + Core Reason: $- \frac{1}{{1 + x^2}}$ . The derivative of the inverse tangent function is used.

Step-by-Step Solution:

Step 1: Recognize that $\tan^{-1}(\frac{{a - x}}{{1 + ax}}) = \tan^{-1}(a) - \tan^{-1}(x)$ using the inverse tangent subtraction formula.

Step 2: Differentiate both sides with respect to $x$ : $\frac{d}{{dx}}\left[ {\tan^{ - 1}\left( {\frac{{a - x}}{{1 + ax}}} \right)} \right] = \frac{d}{{dx}}\left[ {\tan^{ - 1}a - \tan^{ - 1}x} \right]$

Step 3: Apply the derivative rules: $\frac{d}{{dx}}(\tan^{-1} a) = 0$ (since $a$ is a constant) and $\frac{d}{{dx}}(\tan^{-1} x) = \frac{1}{{1 + x^2}}$

Step 4: Substitute the derivatives back into the equation: $\frac{d}{{dx}}\left[ {\tan^{ - 1}\left( {\frac{{a - x}}{{1 + ax}}} \right)} \right] = 0 - \frac{1}{{1 + x^2}} = - \frac{1}{{1 + x^2}}$

Why other options are wrong:
- **Option B ( $\frac{1}{{1 + a^2}} - \frac{1}{{1 + x^2}}$ ):** This is incorrect because it includes a term related to the derivative of $\tan^{-1}(a)$ , which is zero.
- **Option C ( $\frac{1}{{1 + {\left( {\frac{{a - x}}{{1 + ax}}} \right)}^2}}$ ):** This is incorrect because it is the derivative of $\tan^{-1}(\frac{{a - x}}{{1 + ax}})$ without simplification.
- **Option D ( $\frac{{ - 1}}{{\sqrt {1 - {{\left( {\frac{{a - x}}{{1 + ax}}} \right)}^2}} }}$ ):** This is incorrect because it resembles the derivative of $\sin^{-1}(\frac{{a - x}}{{1 + ax}})$ , not $\tan^{-1}(\frac{{a - x}}{{1 + ax}})$ .

Key Concepts:
- Inverse trigonometric functions
- Differentiation of inverse trigonometric functions
- Inverse tangent subtraction formula: $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}(\frac{{A - B}}{{1 + AB}})$

Memory Aids:
- Remember the derivative of $\tan^{-1} x$ is $\frac{1}{{1 + x^2}}$

Quick Tips:
- Use trigonometric identities to simplify the expression before differentiating.
- Recognize common derivative formulas.

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Application +4 | -1

If

f\left( x \right) = \left\{ {\frac{\pi }{2}\left[ x \right] - {x^5}} \right\},1 < x < 2

and [.] denotes the greatest integer function, then

f'\left( {\sqrt[5]{{\frac{\pi }{2}}}} \right)

is equal to

View answer and explanation

Correct option: C

- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}

Option C (

- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}

) is correct because after evaluating the derivative at the given point, the result is

- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}

.

Solution approach

Correct Answer + Core Reason: $- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}$ . After evaluating the derivative at the given point, the result is $- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}$ .

Step-by-Step Solution:

Step 1: Determine the value of [x] in the neighborhood of $x = \sqrt[5]{{\frac{\pi }{2}}}$ .\nSince $1 < x < 2$ , and $x$ is close to $\sqrt[5]{{\frac{\pi }{2}}}$ , $[x] = 1$ .

Step 2: Simplify f(x) in the neighborhood of $x = \sqrt[5]{{\frac{\pi }{2}}}$ .
$f(x) = \sin \left\{ {\frac{\pi }{2}[x] - {x^5}} \right\} = \sin \left( {\frac{\pi }{2} - {x^5}} \right) = \cos {x^5}$

Step 3: Differentiate f(x) with respect to x.
$f'(x) = - 5{x^4}\sin {x^5}$

Step 4: Evaluate f'(x) at $x = \sqrt[5]{{\frac{\pi }{2}}}$ .
$f'\left( {\sqrt[5]{{\frac{\pi }{2}}}} \right) = - 5{\left( {\frac{\pi }{2}} \right)^{\frac{4}{5}}}\sin \frac{\pi }{2} = - 5{\left( {\frac{\pi }{2}} \right)^{\frac{4}{5}}}$

Why other options are wrong:
- Option A: 0 is incorrect because it is not the correct value of the derivative.
- Option B: $5{\left( {\pi /2} \right)^{\frac{4}{5}}}$ is incorrect because it has the wrong sign.
- Option D: None of these is incorrect because the correct answer is $- 5{\left( {\pi /2} \right)^{\frac{4}{5}}}$ .

Key Concepts:
- Greatest integer function
- Differentiation of trigonometric functions
- Chain rule

Memory Aids:
- Remember the properties of the greatest integer function.

Quick Tips:
- Simplify the function using the properties of the greatest integer function before differentiating.

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Analyzing +4 | -1

If

y = f\left( x \right)

and

y\cos x + x\cos y = \mathrm{\pi }

, then the value of

f''\left( 0 \right)

is

View answer and explanation

Correct option: A

\pi

Option A (

\pi

) is correct because, after implicit differentiation and substitution, the value of

f''(0)

is found to be equal to

\pi

.

Solution approach

Correct Answer + Core Reason: $\pi$ . The value of $f''(0)$ is found using implicit differentiation and substitution.

Step-by-Step Solution:

Step 1: Given: $y\cos x + x\cos y = \pi$

Step 2: Differentiate with respect to $x$ using implicit differentiation:
$\frac{{dy}}{{dx}}\cos x - y\sin x + \cos y - x\sin y\frac{{dy}}{{dx}} = 0$

Step 3: Rearrange to solve for $\frac{{dy}}{{dx}}$ :
$\frac{{dy}}{{dx}}\left( {\cos x - x\sin y} \right) = y\sin x - \cos y$
$\frac{{dy}}{{dx}} = \frac{{y\sin x - \cos y}}{{\cos x - x\sin y}}$

Step 4: Differentiate again with respect to $x$ :
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\left( {\cos x - x\sin y} \right)\left( {\frac{{dy}}{{dx}}\sin x + y\cos x + \frac{{dy}}{{dx}}\sin y} \right) - \left( {y\sin x - \cos y} \right)\left( { - \sin x - x\cos y\frac{{dy}}{{dx}} - \sin y} \right)}}{{{{\left( {\cos x - x\sin y} \right)}^2}}}$

Step 5: Evaluate at $x = 0$ :\nFrom the original equation, when $x = 0$ , $y\cos 0 + 0\cos y = \pi \Rightarrow y = \pi$ \nAlso, $\frac{{dy}}{{dx}}\left| {_{x = 0}} \right. = \frac{{\pi \sin 0 - \cos \pi }}{{\cos 0 - 0\sin \pi }} = \frac{{0 - \left( { - 1} \right)}}{1} = 1$

Step 6: Substitute $x = 0$ , $y = \pi$ , and $\frac{{dy}}{{dx}} = 1$ into the second derivative expression:
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{\left( {\cos 0 - 0\sin \pi } \right)\left( {1 \cdot \sin 0 + \pi \cos 0 + 1 \cdot \sin \pi } \right) - \left( {\pi \sin 0 - \cos \pi } \right)\left( { - \sin 0 - 0 \cdot \cos \pi \cdot 1 - \sin \pi } \right)}}{{{{\left( {\cos 0 - 0\sin \pi } \right)}^2}}}$
$\frac{{{d^2}y}}{{d{x^2}}} = \frac{{1\left( {0 + \pi + 0} \right) - \left( {0 - \left( { - 1} \right)} \right)\left( {0 - 0 - 0} \right)}}{{{1^2}}} = \frac{\pi }{1} = \pi$

Why other options are wrong:
- Option B: Incorrect calculation of the second derivative.
- Option C: Incorrect calculation of the second derivative.
- Option D: Incorrect calculation of the second derivative.

Key Concepts:
- Implicit Differentiation

Memory Aids:
- Remember to apply the chain rule when differentiating implicitly.

Quick Tips:
- Evaluate the derivatives at the given point after each differentiation step to simplify the calculations.

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Application +4 | -1

If

y = \frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}}

, then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: D

- \mathrm{cosec}{\mathrm{h}^2}x

Option D (

- \mathrm{cosec}{\mathrm{h}^2}x

) is correct because the derivative of

\coth x

is

- \mathrm{cosec}{\mathrm{h}^2}x

.

Solution approach

Correct Answer + Core Reason: $- \mathrm{cosec}{\mathrm{h}^2}x$ . The derivative of $\coth x$ is $- \mathrm{cosec}{\mathrm{h}^2}x$ .

Step-by-Step Solution:

Step 1: Rewrite the function in terms of hyperbolic functions.
$y = \frac{{{e^x} + {e^{ - x}}}}{{{e^x} - {e^{ - x}}}} = \frac{{\frac{{\left( {{e^x} + {e^{ - x}}} \right)}}{2}}}{{\frac{{\left( {{e^x} - {e^{ - x}}} \right)}}{2}}} = \frac{{\cosh x}}{{\sinh x}}$

Step 2: Simplify the expression.
$y = \coth x$

Step 3: Differentiate with respect to x.
$\frac{{dy}}{{dx}} = - \mathrm{cosec}{\mathrm{h}^2}x$

Why other options are wrong:
- Option A: $\mathrm{sec}{\mathrm{h}^2}x$ is the derivative of $\tanh x$ , not $\coth x$ .
- Option B: $\mathrm{cosec}{\mathrm{h}^2}x$ is the negative of the correct derivative.
- Option C: $- \mathrm{sec}{\mathrm{h}^2}x$ is the negative of the derivative of $\tanh x$ .

Key Concepts:
- Hyperbolic functions
- Differentiation of hyperbolic functions
- $\coth x = \frac{{\cosh x}}{{\sinh x}}$
- $\frac{d}{{dx}}(\coth x) = - \mathrm{cosec}{\mathrm{h}^2}x$

Memory Aids:
- Remember the derivatives of hyperbolic functions.

Quick Tips:
- Convert exponential forms to hyperbolic functions to simplify differentiation.

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Analyzing +4 | -1

Let

f\left( {x + y} \right) = f\left( x \right)f\left( y \right)

and

f\left( x \right) = 1 + \left( {\sin 2x} \right)g\left( x \right)

where

g\left( x \right)

is continuous. Then,

f'\left( x \right)

equals

View answer and explanation

Correct option: B

2f\left( x \right)g\left( 0 \right)

Option B (

2f\left( x \right)g\left( 0 \right)

) is correct because it is the result of applying the limit definition of the derivative and using the given functional equation.

Solution approach

Correct Answer + Core Reason: $2f(x)g(0)$ . This is the derivative obtained using the limit definition and the given functional equation.

Step-by-Step Solution:

Step 1: Use the definition of the derivative: $f'(x) = \lim\limits_{h \to 0} \frac{{f(x + h) - f(x)}}{h}$ .

Step 2: Apply the given functional equation $f(x + y) = f(x)f(y)$ : $f'(x) = \lim\limits_{h \to 0} \frac{{f(x)f(h) - f(x)}}{h} = f(x)\lim\limits_{h \to 0} \frac{{f(h) - 1}}{h}$ .

Step 3: Use the given expression for $f(x)$ : $f(x) = 1 + (\sin 2x)g(x)$ . So, $f(h) = 1 + (\sin 2h)g(h)$ .

Step 4: Substitute this into the limit: $f'(x) = f(x)\lim\limits_{h \to 0} \frac{{1 + (\sin 2h)g(h) - 1}}{h} = f(x)\lim\limits_{h \to 0} \frac{{(\sin 2h)g(h)}}{h}$ .

Step 5: Rewrite the limit: $f'(x) = f(x)\lim\limits_{h \to 0} \frac{{\sin 2h}}{h} \cdot \lim\limits_{h \to 0} g(h)$ .

Step 6: Evaluate the limits: $\lim\limits_{h \to 0} \frac{{\sin 2h}}{h} = 2$ and $\lim\limits_{h \to 0} g(h) = g(0)$ (since $g(x)$ is continuous).

Step 7: Therefore, $f'(x) = f(x) \cdot 2 \cdot g(0) = 2f(x)g(0)$ .

Why other options are wrong:
- Option A: $f(x)g(0)$ is incorrect because it misses the factor of 2 from the limit of $\frac{{\sin 2h}}{h}$ .
- Option C: $2g(0)$ is incorrect because it does not include the $f(x)$ term.
- Option D: None of these is incorrect because option B is correct.

Key Concepts:
- Definition of the derivative
- Functional equations
- Limits of trigonometric functions
- Continuity

Memory Aids:
- Remember the limit $\lim\limits_{x \to 0} \frac{{\sin x}}{x} = 1$ .

Quick Tips:
- When dealing with functional equations, try to use the given equation to simplify the expression.

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Analyzing +4 | -1

If

\frac{{{d^2}x}}{{d{y^2}}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + \frac{{{d^2}y}}{{d{x^2}}} = k

, then

k

is equal to

View answer and explanation

Correct option: A

0

Option A (0) is correct because the given expression simplifies to zero using the relationship between first and second derivatives.

Solution approach

Correct Answer + Core Reason: 0. The given expression simplifies to 0 using derivative relationships.

Step-by-Step Solution:

Step 1: Given equation: $\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + \frac{{{{{d^2}y}}}}{{d{{{x^2}}}}} = k$

Step 2: Use the relationship $\frac{{dy}}{{dx}} = {\left( {\frac{{dx}}{{dy}}} \right)^{ - 1}}$

Step 3: Find $\frac{{{{{d^2}y}}}}{{d{{{x^2}}}}}$ in terms of $\frac{{dx}}{{dy}}$ and $\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}$ :
$\frac{{{{{d^2}y}}}}{{d{{{x^2}}}}} = - {\left( {\frac{{dx}}{{dy}}} \right)^{ - 2}}\left[ {\frac{d}{{dx}}\left( {\frac{{dx}}{{dy}}} \right)} \right] = - {\left( {\frac{{dx}}{{dy}}} \right)^{ - 2}}\left[ {\frac{d}{{dx}}\left( {\frac{{dx}}{{dy}}} \right)\frac{{dy}}{{dx}}} \right] = - {\left( {\frac{{dy}}{{dx}}} \right)^2}\left[ {\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}.\frac{{dy}}{{dx}}} \right] = - {\left( {\frac{{dy}}{{dx}}} \right)^3}\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}$

Step 4: Substitute $\frac{{{{{d^2}y}}}}{{d{{{x^2}}}}}$ into the given equation:
$\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}{\left( {\frac{{dy}}{{dx}}} \right)^3} + \frac{{{{{d^2}y}}}}{{d{{{x^2}}}}} = \frac{{{{{d^2}x}}}}{{d{{{y^2}}}}}{\left( {\frac{{dy}}{{dx}}} \right)^3} - {\left( {\frac{{dy}}{{dx}}} \right)^3}\frac{{{{{d^2}x}}}}{{d{{{y^2}}}}} = 0$

Final Answer: 0

Why other options are wrong:
- Option B (1): Incorrect. The expression does not simplify to 1.
- Option C (2): Incorrect. The expression does not simplify to 2.
- Option D (None of these): Incorrect. The expression simplifies to 0.

Key Concepts:
- Derivatives
- Second-order derivatives
- Relationship between derivatives of inverse functions

Memory Aids:
- Remember the relationship between $\frac{{dy}}{{dx}}$ and $\frac{{dx}}{{dy}}$ .

Quick Tips:
- Express all derivatives in terms of either $x$ or $y$ before simplifying.
- Be careful with the chain rule.

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Application +4 | -1

If

f'\left( x \right) = \sin (\log x)

and

y = f\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)

, then

\frac{{dy}}{{dx}}

at

x = 1

is equal to

View answer and explanation

Correct option: C

12\sin \log \left( 5 \right)

Option C (

12\sin \log ( 5 )

) is correct because it is the result of applying the chain rule and evaluating the derivative at x=1.

Solution approach

Correct Answer + Core Reason: $12\sin \log \left( 5 \right)$ . The derivative is calculated using the chain rule and evaluated at x=1.

Step-by-Step Solution:

Step 1: Identify the functions: $y = f\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)$ and $f'\left( x \right) = \sin (\log x)$ .

Step 2: Apply the chain rule: $\frac{{dy}}{{dx}} = f'\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)\frac{d}{{dx}}\left( {\frac{{2x + 3}}{{3 - 2x}}} \right)$ .

Step 3: Calculate the derivative of the inner function: $\frac{d}{{dx}}\left( {\frac{{2x + 3}}{{3 - 2x}}} \right) = \frac{{\left( {3 - 2x} \right)\left( 2 \right) - \left( {2x + 3} \right)\left( { - 2} \right)}}{{{{\left( {3 - 2x} \right)}^2}}} = \frac{{6 - 4x + 4x + 6}}{{{{\left( {3 - 2x} \right)}^2}}} = \frac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}$ .

Step 4: Substitute into the chain rule: $\frac{{dy}}{{dx}} = \sin \left[ {\log \left( {\frac{{2x + 3}}{{3 - 2x}}} \right)} \right]\left( {\frac{{12}}{{{{\left( {3 - 2x} \right)}^2}}}} \right)$ .

Step 5: Evaluate at $x = 1$ : $\frac{{dy}}{{dx}}\left|_{x = 1} \right. = \frac{{12}}{{{{\left( {3 - 2(1)} \right)}^2}}}\sin \left[ {\log \left( {\frac{{2(1) + 3}}{{3 - 2(1)}}} \right)} \right] = \frac{{12}}{1}\sin \log (5) = 12\sin \log (5)$ .

Why other options are wrong:
- Option A: This option is incorrect because it does not account for the correct derivative of the inner function.
- Option B: This option is incorrect because it does not account for the correct derivative of the inner function and has incorrect values.
- Option D: This option is incorrect because it does not account for the correct derivative of the inner function and has incorrect values.

Key Concepts:
- Chain Rule
- Differentiation of rational functions
- Logarithmic functions

Memory Aids:
- Remember the chain rule: $\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}$

Quick Tips:
- Always simplify the expression after each differentiation step.
- Double-check the derivative of the inner function.

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Application +4 | -1

If

x = {e^t}\sin t,y = {e^t}\cos t

, then

\frac{{{d^2}y}}{{d{x^2}}}

at

x = \pi

, is

View answer and explanation

Correct option: D

\frac{2}{{{e^\pi }}}

Option D (

\frac{2}{{{e^\pi }}}

) is correct because after finding the first and second derivatives and substituting

t = \pi

, we get

\frac{{{d^2}y}}{{d{x^2}}} = \frac{2}{{{e^\pi }}}

.

Solution approach

Correct Answer + Core Reason: $\frac{2}{{{e^\pi }}}$ . The second derivative is calculated using parametric differentiation.

Step-by-Step Solution:

Step 1: Given $x = {e^t}\sin t$ and $y = {e^t}\cos t$ .

Step 2: Find $\frac{{dx}}{{dt}}$ and $\frac{{dy}}{{dt}}$ .
$\frac{{dx}}{{dt}} = {e^t}\sin t + {e^t}\cos t = {e^t}(\sin t + \cos t)$
$\frac{{dy}}{{dt}} = {e^t}\cos t - {e^t}\sin t = {e^t}(\cos t - \sin t)$

Step 3: Find $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}} = \frac{{{e^t}(\cos t - \sin t)}}{{{e^t}(\sin t + \cos t)}} = \frac{{\cos t - \sin t}}{{\sin t + \cos t}}$ .

Step 4: Find $\frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}(\frac{{dy}}{{dx}}) = \frac{d}{{dt}}(\frac{{dy}}{{dx}})\frac{{dt}}{{dx}} = \frac{d}{{dt}}(\frac{{\cos t - \sin t}}{{\sin t + \cos t}})\frac{1}{{\frac{{dx}}{{dt}}}}$ .

Step 5: Calculate $\frac{d}{{dt}}(\frac{{\cos t - \sin t}}{{\sin t + \cos t}}) = \frac{{(\sin t + \cos t)( - \sin t - \cos t) - (\cos t - \sin t)(\cos t - \sin t)}}{{{{(\sin t + \cos t)}^2}}} = \frac{{ - {{(\sin t + \cos t)}^2} - {{(\cos t - \sin t)}^2}}}{{{{(\sin t + \cos t)}^2}}} = \frac{{ - 2}}{{{{(\sin t + \cos t)}^2}}}$ .

Step 6: Substitute back into the expression for $\frac{{{d^2}y}}{{d{x^2}}}$ : $\frac{{{d^2}y}}{{d{x^2}}} = \frac{{ - 2}}{{{{(\sin t + \cos t)}^2}}} \cdot \frac{1}{{{e^t}(\sin t + \cos t)}} = \frac{{ - 2}}{{{e^t}{{(\sin t + \cos t)}^3}}}$ .

Step 7: Evaluate at $t = \pi$ . Since $x = e^t \sin t$ , when $x = \pi$ , $e^t \sin t = \pi$ . However, the question asks for the value at $x=\pi$ , which means we need to find the corresponding value of $t$ . Since $\sin(\pi) = 0$ , we cannot directly substitute $t = \pi$ . The problem statement is slightly ambiguous. Assuming the question meant to evaluate at $t=\pi$ :
$\frac{{{d^2}y}}{{d{x^2}}}\Big|_{t = \pi} = \frac{{ - 2}}{{{e^\pi}{{(\sin \pi + \cos \pi)}^3}}} = \frac{{ - 2}}{{{e^\pi}{{(0 - 1)}^3}}} = \frac{{ - 2}}{{ - {e^\pi}}} = \frac{2}{{{e^\pi}}}$ .

Why other options are wrong:
- **Option A ( $2{e^\pi }$ ): Incorrect. This is likely due to an error in calculating the derivatives or substituting the value of t.
- Option B ( $\frac{1}{2}{e^\pi }$ ): Incorrect. This is likely due to an error in calculating the derivatives or substituting the value of t.
- Option C ( $\frac{1}{{2{e^\pi }}}$ ): Incorrect. This is likely due to an error in calculating the derivatives or substituting the value of t.

Key Concepts:
- Parametric differentiation
- Second derivative
- Chain rule

Memory Aids:
- Remember the chain rule for differentiation.

Quick Tips:**
- Be careful with signs when differentiating trigonometric functions.
- Remember to use the chain rule when finding the second derivative.

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Application +4 | -1

If

f\left( x \right) = {\log _x}({\log _e}x),

then

f'\left( x \right)

at

x = e

is equal to

View answer and explanation

Correct option: D

\frac{1}{e}

Option D (

\frac{1}{e}

) is correct because after differentiating the given function and substituting x = e, the result simplifies to

\frac{1}{e}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{e}$ . The derivative is calculated and evaluated at x=e.

Step-by-Step Solution:

Step 1: Rewrite the function using the change of base formula for logarithms: $f(x) = \log_x(\log_e x) = \frac{{\log_e(\log_e x)}}{{\log_e x}}$

Step 2: Differentiate $f(x)$ with respect to $x$ using the quotient rule: $f'(x) = \frac{{(\log_e x) \cdot \frac{{d}}{{dx}}(\log_e(\log_e x)) - (\log_e(\log_e x)) \cdot \frac{{d}}{{dx}}(\log_e x)}}{{(\log_e x)^2}}$

Step 3: Apply the chain rule to find the derivatives: $\frac{{d}}{{dx}}(\log_e(\log_e x)) = \frac{1}{{\log_e x}} \cdot \frac{1}{x}$ and $\frac{{d}}{{dx}}(\log_e x) = \frac{1}{x}$

Step 4: Substitute the derivatives back into the expression for $f'(x)$ : $f'(x) = \frac{{(\log_e x) \cdot \frac{1}{{\log_e x}} \cdot \frac{1}{x} - (\log_e(\log_e x)) \cdot \frac{1}{x}}}{{(\log_e x)^2}} = \frac{{\frac{1}{x} - \frac{{\log_e(\log_e x)}}{x}}}{{(\log_e x)^2}} = \frac{{1 - \log_e(\log_e x)}}{{x(\log_e x)^2}}$

Step 5: Evaluate $f'(x)$ at $x = e$ : $f'(e) = \frac{{1 - \log_e(\log_e e)}}{{e(\log_e e)^2}} = \frac{{1 - \log_e(1)}}{{e(1)^2}} = \frac{{1 - 0}}{e} = \frac{1}{e}$

Why other options are wrong:
- Option A (1): This is incorrect because it doesn't account for the logarithmic differentiation correctly.
- Option B (2): This is incorrect because it is not the correct derivative evaluated at x=e.
- Option C (0): This is incorrect because it is not the correct derivative evaluated at x=e.

Key Concepts:
- Differentiation of logarithmic functions
- Quotient rule
- Chain rule

Memory Aids:
- Remember the derivative of $\log_e x$ is $\frac{1}{x}$
- Remember the change of base formula for logarithms

Quick Tips:
- Simplify the function before differentiating if possible.
- Be careful with the chain rule when differentiating composite functions.

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Application +4 | -1

If

y=\log _e\left(\log _e x\right)

, then

e^y \frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

\frac{1}{x}

Option B (1/x) is correct because after applying the chain rule and simplifying, the expression e^y * dy/dx simplifies to 1/x.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{x}}$ . The expression is simplified using the chain rule and properties of logarithms.

Step-by-Step Solution:

Step 1: Write the given function: $y = \log_e (\log_e x)$ .

Step 2: Differentiate $y$ with respect to $x$ using the chain rule.
$\frac{{dy}}{{dx}} = \frac{{1}}{{\log_e x}} \cdot \frac{{1}}{{x}} = \frac{{1}}{{x \log_e x}}$

Step 3: Multiply both sides by $e^y$ .
$e^y \frac{{dy}}{{dx}} = \frac{{e^y}}{{x \log_e x}}$

Step 4: Substitute $e^y = \log_e x$ .
$e^y \frac{{dy}}{{dx}} = \frac{{\log_e x}}{{x \log_e x}}$

Step 5: Simplify the expression.
$e^y \frac{{dy}}{{dx}} = \frac{{1}}{{x}}$

Why other options are wrong:
- Option A: $e^y$ : This is incorrect because it doesn't account for the derivative of the inner logarithm.
- Option C: $\frac{{1}}{{\log x}}$ : This is incorrect because it doesn't account for the derivative of the outer logarithm.
- Option D: $\frac{{1}}{{x \log x}}$ : This is the value of $\frac{{dy}}{{dx}}$ , not $e^y \frac{{dy}}{{dx}}$ .

Key Concepts:
- Chain rule
- Differentiation of logarithmic functions

Memory Aids:
- Remember the chain rule: $\frac{{d}}{{dx}} [f(g(x))] = f'(g(x)) \cdot g'(x)$ .

Quick Tips:
- Always use the chain rule when differentiating composite functions.
- Remember the derivative of $\log_e x$ is $\frac{{1}}{{x}}$ .

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Application +4 | -1

The derivative of

\frac{1+\frac{1}{x}}{1-\frac{1}{x}}

is 

View answer and explanation

Correct option: B

\frac{-2}{(1-x)^2}

Option B (

\frac{-2}{(1-x)^2}

) is correct because simplifying the function and then differentiating yields this result.

Solution approach

Correct Answer + Core Reason: $\frac{-2}{(1-x)^2}$ . The derivative is found by simplifying the function and then differentiating.

Step-by-Step Solution:

Step 1: Write the given function: $y = \frac{1 + \frac{1}{x}}{1 - \frac{1}{x}}$

Step 2: Simplify the function:
$y = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$

Step 3: Differentiate $y$ with respect to $x$ using the quotient rule: $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$ \nLet $u = x+1$ and $v = x-1$ \nThen $u' = 1$ and $v' = 1$

Step 4: Substitute into the quotient rule formula:
$\frac{dy}{dx} = \frac{(1)(x-1) - (x+1)(1)}{(x-1)^2} = \frac{x-1 - x - 1}{(x-1)^2} = \frac{-2}{(x-1)^2}$

Why other options are wrong:
- **Option A ( $\frac{2}{(1+x)^2}$ ):** This is incorrect because the numerator should be -2 and the denominator should be $(x-1)^2$ .
- **Option C ( $\frac{-1}{(1-x)^2}$ ): This is incorrect because the numerator should be -2.
- Option D ( $\frac{3}{(1-x)^2}$ ): This is incorrect because the numerator should be -2.

Key Concepts:
- Derivatives
- Quotient rule of differentiation
- Algebraic simplification

Memory Aids:**
- Remember the quotient rule: $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$

Quick Tips:
- Simplify the function before differentiating to make the calculation easier.

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Application +4 | -1

If

y=f\left(\frac{2 x-1}{x^2+1}\right)

and

f^{\prime}(x)=\sin x^2

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left(\frac{2+2 x-2 x^2}{\left(x^2+1\right)^2}\right)

Option B (

\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left(\frac{2+2 x-2 x^2}{\left(x^2+1\right)^2}\right)

) is correct because it is the result of applying the chain rule to differentiate the composite function.

Solution approach

Correct Answer + Core Reason: $\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left(\frac{2+2 x-2 x^2}{\left(x^2+1\right)^2}\right)$ . The derivative is found by using the chain rule.

Step-by-Step Solution:

Step 1: Write the given function: $y = f\left(\frac{2x-1}{x^2+1}\right)$ and $f'(x) = \sin x^2$

Step 2: Apply the chain rule: $\frac{dy}{dx} = f'\left(\frac{2x-1}{x^2+1}\right) \cdot \frac{d}{dx}\left(\frac{2x-1}{x^2+1}\right)$

Step 3: Find the derivative of $\frac{2x-1}{x^2+1}$ using the quotient rule:
$\frac{d}{dx}\left(\frac{2x-1}{x^2+1}\right) = \frac{(2)(x^2+1) - (2x-1)(2x)}{(x^2+1)^2} = \frac{2x^2 + 2 - 4x^2 + 2x}{(x^2+1)^2} = \frac{-2x^2 + 2x + 2}{(x^2+1)^2}$

Step 4: Substitute $f'\left(\frac{2x-1}{x^2+1}\right) = \sin \left(\frac{2x-1}{x^2+1}\right)^2$ and the derivative from Step 3 into the chain rule formula:
$\frac{dy}{dx} = \sin \left(\frac{2x-1}{x^2+1}\right)^2 \cdot \frac{-2x^2 + 2x + 2}{(x^2+1)^2} = \sin \left(\frac{2x-1}{x^2+1}\right)^2 \left(\frac{2 + 2x - 2x^2}{(x^2+1)^2}\right)$

Why other options are wrong:
- **Option A ( $\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left(\frac{2+2 x+x^2}{\left(x^2+1\right)^2}\right)$ ): This is incorrect because the derivative of the inner function is calculated incorrectly.
- Option C ( $\sin \left(\frac{2 x-1}{x^2+1}\right)^2\left(\frac{2+2 x-x^2}{\left(x^2+1\right)^2}\right)$ ): This is incorrect because the derivative of the inner function is calculated incorrectly.
- Option D (None of the above): This is incorrect because option B is the correct answer.

Key Concepts:
- Derivatives
- Chain rule of differentiation
- Quotient rule of differentiation

Memory Aids:**
- Remember the chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$

Quick Tips:
- Apply the chain rule carefully when differentiating composite functions.

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Application +4 | -1

\frac{d^2 x}{d y^2}

equals

View answer and explanation

Correct option: D

-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}

Option D (- (d²y/dx²) * (dy/dx)^-3) is correct because it is derived from the chain rule and inverse function differentiation.

Solution approach

Correct Answer + Core Reason: $-\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-3}$ . This is the correct expression for the second derivative.

Step-by-Step Solution:

Step 1: Start with $\frac{dx}{dy} = \left( \frac{dy}{dx} \right)^{-1}$ .

Step 2: Differentiate $\frac{dx}{dy}$ with respect to $y$ :
$\frac{d^2 x}{dy^2} = \frac{d}{dy} \left( \frac{dx}{dy} \right) = \frac{d}{dy} \left( \left( \frac{dy}{dx} \right)^{-1} \right)$ .

Step 3: Use the chain rule to rewrite the derivative with respect to $x$ :
$\frac{d^2 x}{dy^2} = \frac{d}{dx} \left( \left( \frac{dy}{dx} \right)^{-1} \right) \frac{dx}{dy}$ .

Step 4: Differentiate $\left( \frac{dy}{dx} \right)^{-1}$ with respect to $x$ :
$\frac{d^2 x}{dy^2} = -\left( \frac{dy}{dx} \right)^{-2} \frac{d^2 y}{dx^2} \frac{dx}{dy}$ .

Step 5: Substitute $\frac{dx}{dy} = \left( \frac{dy}{dx} \right)^{-1}$ :
$\frac{d^2 x}{dy^2} = -\left( \frac{dy}{dx} \right)^{-2} \frac{d^2 y}{dx^2} \left( \frac{dy}{dx} \right)^{-1} = -\left( \frac{d^2 y}{dx^2} \right) \left( \frac{dy}{dx} \right)^{-3}$ .

Why other options are wrong:
- Option A: $\left(\frac{d^2 y}{dx^2}\right)^{-1}$ is incorrect because it only considers the inverse of the second derivative and doesn't account for the chain rule.
- Option B: $-\left(\frac{d^2 y}{d x^2}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}$ is incorrect because the first term should not be inverted.
- Option C: $\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d y}{d x}\right)^{-2}$ is incorrect because the power of $\left(\frac{dy}{dx}\right)$ should be -3, not -2, and the sign is incorrect.

Key Concepts:
- Chain rule
- Inverse function differentiation
- Second derivative

Memory Aids:
- Remember the chain rule: $\frac{d}{dy}f(x) = \frac{d}{dx}f(x) \frac{dx}{dy}$ .

Quick Tips:
- When finding higher-order derivatives of inverse functions, use the chain rule carefully.
- Keep track of which variable you are differentiating with respect to.

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Application +4 | -1

If

y=\log _a x+\log _x a+\log _x x+\log _a a

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: D

\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}

Option D (

\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}

) is correct because after applying the change of base formula for logarithms and differentiating, the derivative is found to be

\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}$ . The derivative is found by applying logarithm rules and differentiating.

Step-by-Step Solution:

Step 1: Apply the change of base formula to the logarithms.
$y = \log_a x + \log_x a + \log_x x + \log_a a = \frac{\log x}{\log a} + \frac{\log a}{\log x} + 1 + 1 = \frac{\log x}{\log a} + \frac{\log a}{\log x} + 2$

Step 2: Differentiate with respect to $x$ .
$\frac{dy}{dx} = \frac{d}{dx}(\frac{\log x}{\log a} + \frac{\log a}{\log x} + 2) = \frac{1}{\log a} \frac{d}{dx}(\log x) + \log a \frac{d}{dx}(\frac{1}{\log x}) + 0 = \frac{1}{\log a} (\frac{1}{x}) + \log a (-\frac{1}{(\log x)^2} \frac{1}{x})$

Step 3: Simplify the expression.
$\frac{dy}{dx} = \frac{1}{x \log a} - \frac{\log a}{x(\log x)^2}$

Why other options are wrong:
- **Option A ( $\frac{1}{x}+x \log a$ ): This is incorrect because the logarithm rules and differentiation were not applied correctly.
- Option B ( $\frac{\log a}{x}+\frac{x}{\log a}$ ): This is incorrect because the logarithm rules and differentiation were not applied correctly.
- Option C ( $\frac{1}{x \log a}+x \log a$ ): This is incorrect because the logarithm rules and differentiation were not applied correctly. The sign is also incorrect.

Key Concepts:**
- Change of base formula for logarithms: $\log_a x = \frac{\log x}{\log a}$
- Differentiation of logarithms

Memory Aids:
- Remember the change of base formula for logarithms.
- Recall the derivative of $\log x$ is $\frac{1}{x}$ .

Quick Tips:
- Use the change of base formula to simplify logarithmic expressions before differentiating.
- Be careful with the chain rule when differentiating composite functions.

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Application +4 | -1

If function

f(x)

is differentiable at

x=a

, then

\lim _{x \rightarrow a} \frac{x^{2} f(a)-a^{2} f(x)}{x-a}

is :

View answer and explanation

Correct option: C

2 a f(a)-a^{2} f^{\prime}(a)

Option C (

2 a f(a)-a^{2} f^{\prime}(a)

) is correct because it is the result of applying L'Hopital's rule to the given limit.

Solution approach

Correct Answer + Core Reason: $2 a f(a)-a^{2} f^{\prime}(a)$ . This is obtained by using L'Hopital's rule.

Step-by-Step Solution:

Step 1: Write the given limit: $\lim_{x \to a} \frac{x^2 f(a) - a^2 f(x)}{x - a}$ .

Step 2: Check if the limit is in indeterminate form: As $x \to a$ , the numerator approaches $a^2 f(a) - a^2 f(a) = 0$ , and the denominator approaches $a - a = 0$ . So, it is in the indeterminate form $\frac{0}{0}$ .

Step 3: Apply L'Hopital's rule: Differentiate the numerator and denominator with respect to $x$ :
$\frac{d}{dx}(x^2 f(a) - a^2 f(x)) = 2x f(a) - a^2 f'(x)$ .
$\frac{d}{dx}(x - a) = 1$ .

Step 4: Rewrite the limit using L'Hopital's rule: $\lim_{x \to a} \frac{2x f(a) - a^2 f'(x)}{1}$ .

Step 5: Evaluate the limit: As $x \to a$ , the expression becomes $2a f(a) - a^2 f'(a)$ .

Why other options are wrong:
- **Option A ( $-a^{2} f^{\prime}(a)$ ):** This is incorrect because it only considers the derivative term and misses the term involving $f(a)$ .
- **Option B ( $a f(a)-a^{2} f^{\prime}(a)$ ):** This is incorrect because the coefficient of $af(a)$ is wrong. It should be $2af(a)$ .
- **Option D ( $2 a f(a)+a^{2} f^{\prime}(a)$ ):** This is incorrect because the sign of the derivative term is wrong. It should be $-a^2 f'(a)$ .

Key Concepts:
- L'Hopital's rule
- Differentiation
- Limits

Memory Aids:
- L'Hopital's rule: If $\lim_{x \to a} \frac{f(x)}{g(x)}$ is in the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ , then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ , provided the limit exists.

Quick Tips:
- Always check if the limit is in an indeterminate form before applying L'Hopital's rule.
- Remember to differentiate both the numerator and the denominator separately.

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Application +4 | -1

If

y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right) \frac{d y}{d x}=

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because after simplification using trigonometric identities and substitution, the derivative of y with respect to x is -1.

Solution approach

Correct Answer + Core Reason: -1. The derivative is found by simplifying the expression using trigonometric identities and then differentiating.

Step-by-Step Solution:

Step 1: Rewrite the given expression using $a = r \cos \alpha$ and $b = r \sin \alpha$ .
$y = \tan^{-1} \left( \frac{{r \cos \alpha \cos x - r \sin \alpha \sin x}}{{r \sin \alpha \cos x + r \cos \alpha \sin x}} \right)$

Step 2: Simplify the expression inside the arctangent function.
$y = \tan^{-1} \left( \frac{{\cos \alpha \cos x - \sin \alpha \sin x}}{{\sin \alpha \cos x + \cos \alpha \sin x}} \right) = \tan^{-1} \left( \frac{{\cos(x + \alpha)}}{{\sin(x + \alpha)}} \right) = \tan^{-1} \left( \cot(x + \alpha) \right)$

Step 3: Rewrite the cotangent function in terms of tangent.
$y = \tan^{-1} \left( \tan(\frac{\pi}{2} - (x + \alpha)) \right)$

Step 4: Simplify the expression further.
$y = \frac{\pi}{2} - x - \alpha$

Step 5: Differentiate y with respect to x.
$\frac{{dy}}{{dx}} = -1$

Why other options are wrong:
- Option A: $\frac{{1}}{{1 + x^2}}$ : This is the derivative of $\tan^{-1} x$ , not the given function.
- Option B: $\frac{{1}}{{\sqrt{{1 - x^2}}}}$ : This is the derivative of $\sin^{-1} x$ or $\cos^{-1} x$ , not the given function.
- Option D: none of these: This is incorrect because the derivative is -1.

Key Concepts:
- Trigonometric identities
- Inverse trigonometric functions
- Differentiation

Memory Aids:
- $\cos(A + B) = \cos A \cos B - \sin A \sin B$
- $\sin(A + B) = \sin A \cos B + \cos A \sin B$
- $\cot \theta = \frac{{1}}{{\tan \theta}}$
- $\tan^{-1}(\tan x) = x$

Quick Tips:
- Simplify the expression before differentiating to make the process easier.
- Remember trigonometric identities to simplify complex expressions.

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Application +4 | -1

If

y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 m}\right)

, then the value of

\frac{d y}{d x}

at

x=0

is ......

View answer and explanation

Correct option: C

1

Option C is correct because the derivative evaluated at x=0 is 1.

Solution approach

Correct Answer + Core Reason: 1. The derivative is evaluated at x=0.

Step-by-Step Solution:

Step 1: Given equation: $y=(1+x)(1+x^2)(1+x^4) \ldots(1+x^{2m})$

Step 2: Take the logarithm of both sides:
$\log y = \log(1+x) + \log(1+x^2) + \log(1+x^4) + \ldots + \log(1+x^{2m})$

Step 3: Differentiate both sides with respect to x:
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \ldots + \frac{2mx^{2m-1}}{1+x^{2m}}$

Step 4: Multiply both sides by y:
$\frac{dy}{dx} = y \left[ \frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \ldots + \frac{2mx^{2m-1}}{1+x^{2m}} \right]$

Step 5: Evaluate at x=0:\nWhen x=0, y = (1+0)(1+0)(1+0)...(1+0) = 1
$\left( \frac{dy}{dx} \right)_{x=0} = 1 \left[ \frac{1}{1+0} + \frac{2(0)}{1+0} + \frac{4(0)}{1+0} + \ldots + \frac{2m(0)}{1+0} \right] = 1[1+0+0+...+0]$

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Application +4 | -1

If

y=\log _{x^2+4}\left(7 x^2-5 x+1\right)

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: A

\frac{1}{\log _e\left(x^2+4\right)}\left(\frac{14 x-5}{7 x^2-5 x+1}-\frac{2 x y}{x^2+4}\right)

Option A is correct because it is the result of applying the change of base formula and the quotient rule for differentiation.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\log _e(x^2+4)}(\frac{14 x-5}{7 x^2-5 x+1}-\frac{2 x y}{x^2+4})$ . This is the result of applying the change of base formula and the quotient rule.

Step-by-Step Solution:

Step 1: Given $y = \log_{x^2+4}(7x^2 - 5x + 1)$

Step 2: Change of base formula: $y = \frac{\log_e(7x^2 - 5x + 1)}{\log_e(x^2 + 4)}$

Step 3: Let $f(x) = 7x^2 - 5x + 1$ and $g(x) = x^2 + 4$

Step 4: Differentiate using the quotient rule: $\frac{dy}{dx} = \frac{\log_e(g(x)) \frac{f'(x)}{f(x)} - \log_e(f(x)) \frac{g'(x)}{g(x)}}{[\log_e(g(x))]^2}$

Step 5: Simplify: $\frac{dy}{dx} = \frac{1}{\log_e(g(x))} [\frac{f'(x)}{f(x)} - \frac{\log_e(f(x))}{\log_e(g(x))} \frac{g'(x)}{g(x)}] = \frac{1}{\log_e(g(x))} [\frac{f'(x)}{f(x)} - y \frac{g'(x)}{g(x)}]$

Step 6: Find $f'(x) = 14x - 5$ and $g'(x) = 2x$

Step 7: Substitute: $\frac{dy}{dx} = \frac{1}{\log_e(x^2+4)} (\frac{14x - 5}{7x^2 - 5x + 1} - y \frac{2x}{x^2 + 4})$

Why other options are wrong:
- Option B: $\frac{1}{\log _e(x^2+4)}(\frac{14 x-5}{7 x^2-5 x+1}+\frac{2 x y}{x^2+4})$ - The sign between the two terms is incorrect. It should be subtraction, not addition.
- Option C: $-\frac{1}{\log _e(x^2+4)}(\frac{14 x-5}{7 x^2-5 x+1}-\frac{2 x y}{x^2+4})$ - The overall sign is incorrect. It should be positive, not negative.
- Option D: none of these - Option A is the correct answer.

Key Concepts:
- Change of Base Formula for Logarithms
- Quotient Rule for Differentiation

Memory Aids:
- Remember the change of base formula: $\log_a b = \frac{\log_c b}{\log_c a}$

Quick Tips:
- When differentiating logarithmic functions with variable bases, always use the change of base formula first.

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Analyzing +4 | -1

If

f(x)

is continuous and differentiable function such that

f\left(\frac{1}{n}\right)=0

for all

n \in N

, then

View answer and explanation

Correct option: B

f(0)=0, f^{\prime}(0)=0

Option B (

f(0)=0, f^{\prime}(0)=0

) is correct because the given condition

f(\frac{1}{n})=0

for all

n \in N

implies that

f(0)=0

and

f^{\prime}(0)=0

due to continuity and differentiability.

Solution approach

Correct Answer + Core Reason: $f(0)=0, f^{\prime}(0)=0$ . The given condition $f(\frac{1}{n})=0$ for all $n \in N$ implies that $f(0)=0$ and $f^{\prime}(0)=0$ .

Step-by-Step Solution:

Step 1: Given $f(\frac{1}{n}) = 0$ for all $n \in N$ , and $f(x)$ is continuous and differentiable.

Step 2: Since $f$ is continuous, $\lim_{n \to \infty} f(\frac{1}{n}) = f(\lim_{n \to \infty} \frac{1}{n}) = f(0)$ .

Step 3: Since $f(\frac{1}{n}) = 0$ for all $n$ , $\lim_{n \to \infty} f(\frac{1}{n}) = 0$ . Therefore, $f(0) = 0$ .

Step 4: Now, consider the derivative $f^{\prime}(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h}$ .

Step 5: Let $h = \frac{1}{n}$ . Then $f^{\prime}(0) = \lim_{n \to \infty} \frac{f(\frac{1}{n}) - f(0)}{\frac{1}{n}} = \lim_{n \to \infty} \frac{0 - 0}{\frac{1}{n}} = 0$ .

Step 6: Therefore, $f(0) = 0$ and $f^{\prime}(0) = 0$ .

Final Answer: $f(0)=0, f^{\prime}(0)=0$

Why other options are wrong:
- Option A: This is incorrect because it states $f(x) = 0$ for all $x \in N \cup (0, 1]$ , which is not necessarily true. We only know that $f(\frac{1}{n}) = 0$ for $n \in N$ .
- Option C: This is incorrect because it only states $f^{\prime}(0) = 0$ and $f^{\prime \prime}(0) = 0$ , but we have shown that $f(0) = 0$ as well.
- Option D: This is incorrect because we have shown that $f(0)$ and $f^{\prime}(0)$ must be zero.

Key Concepts:
- Continuity
- Differentiability
- Limits

Memory Aids:
- Remember that if a function is continuous, the limit of the function at a point is equal to the function's value at that point.

Quick Tips:
- Use the definition of continuity and differentiability to solve problems involving these concepts.
- When given a condition like $f(\frac{1}{n}) = 0$ , try to take the limit as $n \to \infty$ to find the value of $f(0)$ .

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Application +4 | -1

If

x = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {t^2}} }}} \right),y = {\sin ^{ - 1}}\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right) \Rightarrow \frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: C

1

Option C (1) is correct because both x and y simplify to

tan^{-1}(t)

, making

\frac{{dy}}{{dx}}

equal to 1.

Solution approach

Correct Answer + Core Reason: 1. Both x and y simplify to $tan^{-1}(t)$ , making $\frac{{dy}}{{dx}}$ equal to 1.

Step-by-Step Solution:

Step 1: Simplify x.
$x = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt {1 + {t^2}} }}} \right) = {\tan ^{ - 1}}t$

Step 2: Simplify y.
$y = {\sin ^{ - 1}}\left( {\frac{t}{{\sqrt {1 + {t^2}} }}} \right) = {\tan ^{ - 1}}t$

Step 3: Find $\frac{{dy}}{{dx}}$ .\nSince $x = {\tan ^{ - 1}}t$ and $y = {\tan ^{ - 1}}t$ , $y = x$ . Therefore, $\frac{{dy}}{{dx}} = 1$ .

Why other options are wrong:
- Option A: 0 is incorrect because y=x, so the derivative is 1.
- Option B: $\tan t$ is incorrect because it is not the correct derivative.
- Option D: $\sin t\cos t$ is incorrect because it is not the correct derivative.

Key Concepts:
- Inverse trigonometric functions

Memory Aids:
- Remember the relationships between inverse trigonometric functions.

Quick Tips:
- Simplify the expressions before differentiating.

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Application +4 | -1

\frac{d^{2} x}{d y^{2}}

equals :

View answer and explanation

Correct option: C

-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}

Option C (-(

\frac{{d^{2}y}}{{dx^{2}}}

)(

\frac{{dy}}{{dx}}

)

^{-3}

) is correct because it represents the correct expression for

\frac{{d^{2}x}}{{dy^{2}}}

in terms of derivatives of y with respect to x.

Solution approach

Correct Answer + Core Reason: -( $\frac{{d^{2}y}}{{dx^{2}}}$ )( $\frac{{dy}}{{dx}}$ ) $^{-3}$ . The expression for $\frac{{d^{2}x}}{{dy^{2}}}$ is -( $\frac{{d^{2}y}}{{dx^{2}}}$ )( $\frac{{dy}}{{dx}}$ ) $^{-3}$ .

Step-by-Step Solution:

Step 1: Start with $\frac{{dx}}{{dy}} = \frac{{1}}{{\frac{{dy}}{{dx}}}}$ .

Step 2: Differentiate both sides with respect to $y$ : $\frac{{d^2x}}{{dy^2}} = \frac{{d}}{{dy}} \left( \frac{{1}}{{\frac{{dy}}{{dx}}}} \right)$ .

Step 3: Use the chain rule: $\frac{{d^2x}}{{dy^2}} = \frac{{d}}{{dx}} \left( \frac{{1}}{{\frac{{dy}}{{dx}}}} \right) \cdot \frac{{dx}}{{dy}}$ .

Step 4: Differentiate $\frac{{1}}{{\frac{{dy}}{{dx}}}}$ with respect to $x$ : $\frac{{d}}{{dx}} \left( \frac{{1}}{{\frac{{dy}}{{dx}}}} \right) = - \frac{{\frac{{d^2y}}{{dx^2}}}}{{(\frac{{dy}}{{dx}})^2}}$ .

Step 5: Substitute back into the equation: $\frac{{d^2x}}{{dy^2}} = - \frac{{\frac{{d^2y}}{{dx^2}}}}{{(\frac{{dy}}{{dx}})^2}} \cdot \frac{{1}}{{\frac{{dy}}{{dx}}}} = - \frac{{\frac{{d^2y}}{{dx^2}}}}{{(\frac{{dy}}{{dx}})^3}}$ .

Step 6: Rewrite the expression: $\frac{{d^2x}}{{dy^2}} = - \left( \frac{{d^2y}}{{dx^2}} \right) \left( \frac{{dy}}{{dx}} \right)^{-3}$ .

Why other options are wrong:
- **Option A (-( $\frac{{d^{2}y}}{{dx^{2}}}$ ) $^{-1}$ ( $\frac{{dy}}{{dx}}$ ) $^{-3}$ ): Incorrectly inverts the second derivative.
- Option B ( $\frac{{d^{2}y}}{{dx^{2}}}$ ( $\frac{{dy}}{{dx}}$ ) $^{-2}$ ): Incorrectly uses the power of the derivative.
- Option D (( $\frac{{d^{2}y}}{{dx^{2}}}$ ) $^{-1}$ ): Incorrectly inverts the second derivative and misses the power of the first derivative.

Key Concepts:
- Chain Rule
- Implicit Differentiation
- Second Derivatives

Memory Aids:**
- Remember to use the chain rule when differentiating with respect to a different variable.
- Recall that $\frac{{dx}}{{dy}} = \frac{{1}}{{\frac{{dy}}{{dx}}}}$ .

Quick Tips:
- Always use the chain rule when differentiating composite functions.
- Be careful with the powers of the derivatives when applying the chain rule.

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Application +4 | -1

If

y=a e^{m x}+b e^{-a x}

, then

y_2

is equal to

View answer and explanation

Correct option: B

m^2 y

Option B (

m^2 y

) is correct because differentiating the given equation twice leads to this result.

Solution approach

Correct Answer + Core Reason: $m^2 y$ . The second derivative is found by differentiating the given equation twice.

Step-by-Step Solution:

Step 1: Given $y = a e^{mx} + b e^{-mx}$

Step 2: Find the first derivative $y_1 = \frac{dy}{dx}$ : $y_1 = a e^{mx} m + b e^{-mx}(-m) = m(a e^{mx} - b e^{-mx})$

Step 3: Find the second derivative $y_2 = \frac{d^2y}{dx^2}$ : $y_2 = m(a e^{mx} m - b e^{-mx}(-m)) = m^2(a e^{mx} + b e^{-mx})$

Step 4: Substitute $y = a e^{mx} + b e^{-mx}$ : $y_2 = m^2 y$

Why other options are wrong:
- Option A: $-m^2 y$ - The sign is incorrect. It should be positive.
- Option C: $m y_1$ - This is incorrect because $m y_1 = m^2(a e^{mx} - b e^{-mx})$ , which is not equal to $y_2$ .
- Option D: none of these - Option B is the correct answer.

Key Concepts:
- Differentiation
- Exponential Functions

Memory Aids:
- Remember the derivative of $e^{kx}$ is $ke^{kx}$ .

Quick Tips:
- When dealing with exponential functions, the derivatives often involve the original function.

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Analyzing +4 | -1

The function

f(x)=\left(x^2-1\right)\left|x^2-3 x+2\right|+\cos (|x|)

is not differentiable at

View answer and explanation

Correct option: D

2

Option D (2) is correct because the function is not differentiable at x=2 due to the absolute value function changing its behavior at that point.

Solution approach

Correct Answer + Core Reason: 2. The function is not differentiable at x=2 because the left and right derivatives are not equal.

Step-by-Step Solution:

Step 1: Analyze the absolute value function: $|x^2-3x+2| = |(x-1)(x-2)|$ . This function changes its behavior at x=1 and x=2.

Step 2: Rewrite the function f(x) piecewise:\nf(x) = (x^2-1)|x^2-3x+2| + cos(|x|) =
\begin{{cases}} (x^2-1)(x^2-3x+2) + cos(x), & x < 1 \\ -(x^2-1)(x^2-3x+2) + cos(x), & 1 ≤ x ≤ 2 \\ (x^2-1)(x^2-3x+2) + cos(x), & x > 2 \\ \end{{cases}}

Step 3: Find the derivative of f(x) piecewise:\nf'(x) = \begin{{cases}} (2x)(x^2-3x+2) + (x^2-1)(2x-3) - sin(x), & x < 1 \\ -(2x)(x^2-3x+2) - (x^2-1)(2x-3) - sin(x), & 1 ≤ x ≤ 2 \\ (2x)(x^2-3x+2) + (x^2-1)(2x-3) - sin(x), & x > 2 \\ \end{{cases}}

Step 4: Check differentiability at x=1:\nf'(1^-) = (2)(-sin(1)) = -sin(1)\nf'(1⁺) = -sin(1)\nSince f'(1^-) = f'(1⁺), f(x) is differentiable at x=1.

Step 5: Check differentiability at x=2:\nf'(2^-) = -(4-1)(4-6+2) - (4-1)(4-3) - sin(2) = -3 - sin(2)\nf'(2⁺) = (4-1)(4-6+2) + (4-1)(4-3) - sin(2) = 3 - sin(2)\nSince f'(2^-) != f'(2⁺), f(x) is not differentiable at x=2.

Why other options are wrong:
- Option A (-1): The function is differentiable at x = -1.
- Option B (0): The function is differentiable at x = 0.
- Option C (1): The function is differentiable at x = 1.
- Option D (2): The function is NOT differentiable at x = 2, making this the correct answer.

Key Concepts:
- Differentiability
- Absolute value function
- Piecewise functions
- Derivatives of trigonometric functions

Memory Aids:
- Remember that absolute value functions are often not differentiable at points where the expression inside the absolute value is zero.

Quick Tips:
- When dealing with absolute value functions, rewrite them as piecewise functions to analyze their differentiability.

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Application +4 | -1

\frac{d^{2} x}{d y^{2}}

equals

View answer and explanation

Correct option: C

-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}

The correct answer is C (

-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}

) because it represents the second derivative of x with respect to y in terms of derivatives of y with respect to x.

Solution approach

Correct Answer + Core Reason: $-\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}$ . The second derivative $\frac{d^2x}{dy^2}$ is found by using the inverse relationship between $\frac{dy}{dx}$ and $\frac{dx}{dy}$ .

Step-by-Step Solution:

Step 1: We know that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}} = \left(\frac{dy}{dx}\right)^{-1}$ .

Step 2: Differentiate with respect to $y$ : $\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dy}\left(\left(\frac{dy}{dx}\right)^{-1}\right)$ .

Step 3: Apply the chain rule: $\frac{d^2x}{dy^2} = -1\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d}{dy}\left(\frac{dy}{dx}\right)$ .

Step 4: Rewrite $\frac{d}{dy}\left(\frac{dy}{dx}\right)$ using $\frac{dy}{dx} = \frac{dy}{dx} \cdot \frac{dx}{dx}$ : $\frac{d}{dy}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(\frac{dy}{dx}\right) \cdot \frac{dx}{dy} = \frac{d^2y}{dx^2} \cdot \frac{dx}{dy}$ .

Step 5: Substitute $\frac{dx}{dy} = \left(\frac{dy}{dx}\right)^{-1}$ : $\frac{d^2x}{dy^2} = -\left(\frac{dy}{dx}\right)^{-2} \cdot \frac{d^2y}{dx^2} \cdot \left(\frac{dy}{dx}\right)^{-1}$ .

Step 6: Simplify: $\frac{d^2x}{dy^2} = -\frac{d^2y}{dx^2} \cdot \left(\frac{dy}{dx}\right)^{-3}$ .

Step 7: Final Answer: $\frac{d^2x}{dy^2} = -\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-3}$ .

Why other options are wrong:
- Option A: $-\left(\frac{d^{2} y}{d x^{2}}\right)^{-1}\left(\frac{d y}{d x}\right)^{-3}$ : This is incorrect as it has the inverse of the second derivative of y with respect to x, which is not the correct expression.
- Option B: $\left(\frac{d^{2} y}{d x^{2}}\right)\left(\frac{d y}{d x}\right)^{-2}$ : This is incorrect as it has the wrong power for $\left(\frac{dy}{dx}\right)$ and the wrong sign.
- Option D: $\left(\frac{d^{2} y}{d x^{2}}\right)^{-1}$ : This is incorrect as it only considers the inverse of the second derivative of y with respect to x and ignores the other terms.

Key Concepts:
- Inverse functions
- Chain rule
- Second derivative

Memory Aids:
- Remember that $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$ .
- Use the chain rule carefully when differentiating with respect to different variables.

Quick Tips:
- Express the derivatives in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ .
- Use the chain rule carefully.
- Simplify the expression after each step to avoid errors.

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Application +4 | -1

If

x = a(\theta - \sin \theta ),y = a\left( {1 - \cos \theta } \right),

then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: A

\cot \frac{\mathrm{\theta }}{2}

Option A (

\cot \frac{\mathrm{\theta }}{2}

) is the correct answer because it is the simplified form of

\frac{{dy}}{{dx}}

after differentiating

x

and

y

with respect to

\theta

and applying trigonometric identities.

Solution approach

Correct Answer + Core Reason: $\cot \frac{\mathrm{\theta }}{2}$ . The derivative is found by differentiating x and y with respect to theta and simplifying.

Step-by-Step Solution:

Step 1: Given $x = a(\theta - \sin \theta ), y = a(1 - \cos \theta )$ .

Step 2: Find $\frac{{dx}}{{d\theta }}$ and $\frac{{dy}}{{d\theta }}$ .
$\frac{{dx}}{{d\theta }} = a(1 - \cos \theta )$
$\frac{{dy}}{{d\theta }} = a\sin \theta$

Step 3: Calculate $\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{d\theta }}}}{{\frac{{dx}}{{d\theta }}}}$ .
$\frac{{dy}}{{dx}} = \frac{{a\sin \theta }}{{a(1 - \cos \theta )}}$

Step 4: Simplify using trigonometric identities: $\sin \theta = 2\sin \frac{{\theta }}{2}\cos \frac{{\theta }}{2}$ and $1 - \cos \theta = 2{\sin ^2}\frac{{\theta }}{2}$ .
$\frac{{dy}}{{dx}} = \frac{{2\sin \frac{{\theta }}{2}\cos \frac{{\theta }}{2}}}{{2{{\sin }^2}\frac{{\theta }}{2}}} = \cot \frac{{\theta }}{2}$

Why other options are wrong:
- Option B: $\tan \frac{\mathrm{\theta }}{2}$ is incorrect because it's the reciprocal of the correct answer.
- Option C: $\frac{1}{2}\mathrm{cose}{\mathrm{c}^2}\frac{\mathrm{\theta }}{2}$ is incorrect. It might result from incorrectly differentiating or simplifying.
- Option D: $- \frac{1}{2}\mathrm{cose}{\mathrm{c}^2}\frac{\mathrm{\theta }}{2}$ is incorrect. It might result from incorrectly differentiating or simplifying.

Key Concepts:
- Parametric Differentiation
- Trigonometric Identities

Memory Aids:
- Remember the double angle formulas for sine and cosine.

Quick Tips:
- Always simplify trigonometric expressions after differentiation.

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Application +4 | -1

Let

y=y(x)

be

a

function of

x

satisfying

y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}}

where

k

is

a

constant and

y\left(\frac{1}{2}\right)=-\frac{1}{4}

. Then

\frac{d y}{d x}

at

x=\frac{1}{2}

, is equal to:

View answer and explanation

Correct option: B

-\frac{\sqrt{5}}{2}

Option B (

-\frac{\sqrt{5}}{2}

) is correct because it is the value of

\frac{dy}{dx}

obtained by implicit differentiation and substitution of the given values.

Solution approach

Correct Answer + Core Reason: $-\frac{\sqrt{5}}{2}$ . This is found by implicit differentiation and substitution.

Step-by-Step Solution:

Step 1: Write the given equation: $y \sqrt{1 - x^2} = k - x \sqrt{1 - y^2}$ .

Step 2: Find the value of $k$ using $y(\frac{1}{2}) = -\frac{1}{4}$ : $-\frac{1}{4} \sqrt{1 - (\frac{1}{2})^2} = k - \frac{1}{2} \sqrt{1 - (- \frac{1}{4})^2} \Rightarrow -\frac{1}{4} \sqrt{\frac{3}{4}} = k - \frac{1}{2} \sqrt{\frac{15}{16}} \Rightarrow -\frac{\sqrt{3}}{8} = k - \frac{\sqrt{15}}{8} \Rightarrow k = \frac{\sqrt{15} - \sqrt{3}}{8}$ .

Step 3: Differentiate both sides of the equation with respect to $x$ :
$\frac{d}{dx}(y \sqrt{1 - x^2}) = \frac{d}{dx}(k - x \sqrt{1 - y^2})$ .

Step 4: Apply the product rule and chain rule:
$y' \sqrt{1 - x^2} + y \cdot \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = 0 - [\sqrt{1 - y^2} + x \cdot \frac{1}{2\sqrt{1 - y^2}} \cdot (-2y) \cdot y']$ .

Step 5: Simplify the equation:\ny' $\sqrt{1 - x^2}$ - $\frac{xy}{\sqrt{1 - x^2}}$ = - $\sqrt{1 - y^2}$ + $\frac{xyy'}{\sqrt{1 - y^2}}$ $.\n\n**Step 6:** Substitute$ x = $\frac{1}{2}$ $and$ y = - $\frac{1}{4}$ $:\ny' \sqrt{1 - (\frac{1}{2})^2} - \frac{(\frac{1}{2})(-\frac{1}{4})}{\sqrt{1 - (\frac{1}{2})^2}} = -\sqrt{1 - (- \frac{1}{4})^2} + \frac{(\frac{1}{2})(-\frac{1}{4})y'}{\sqrt{1 - (- \frac{1}{4})^2}}$ .

Step 7: Simplify further:\ny' $\sqrt{\frac{3}{4}}$ + $\frac{1/8}{\sqrt{3/4}}$ = - $\sqrt{\frac{15}{16}}$ - $\frac{y'/8}{\sqrt{15/16}}$ $.\n\n**Step 8:** Solve for$ y' $:\ny' \frac{\sqrt{3}}{2} + \frac{1}{8} \cdot \frac{2}{\sqrt{3}} = -\frac{\sqrt{15}}{4} - \frac{y'}{8} \cdot \frac{4}{\sqrt{15}}\Rightarrow y' \frac{\sqrt{3}}{2} + \frac{1}{4\sqrt{3}} = -\frac{\sqrt{15}}{4} - \frac{y'}{2\sqrt{15}}\Rightarrow y'(\frac{\sqrt{3}}{2} + \frac{1}{2\sqrt{15}}) = -\frac{\sqrt{15}}{4} - \frac{1}{4\sqrt{3}}\Rightarrow y'(\frac{\sqrt{45} + 1}{2\sqrt{15}}) = -\frac{\sqrt{45} + 1}{4\sqrt{3}}\Rightarrow y' = -\frac{2\sqrt{15}}{4\sqrt{3}} = -\frac{\sqrt{5}}{2}$ .

Why other options are wrong:
- **Option A ( $-\frac{\sqrt{5}}{4}$ ):** This is incorrect due to an error in the algebraic manipulation while solving for $y'$ .
- **Option C ( $\frac{2}{\sqrt{5}}$ ): This is incorrect, likely due to a sign error or an incorrect simplification.
- Option D ( $\frac{\sqrt{5}}{2}$ ): This is incorrect because it has the wrong sign. The correct answer is negative.

Key Concepts:
- Implicit differentiation
- Product rule
- Chain rule

Memory Aids:
- Remember to use the product rule when differentiating a product of two functions.
- Remember to use the chain rule when differentiating a composite function.

Quick Tips:**
- After differentiating, substitute the given values to simplify the equation and solve for the derivative.

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Differentiation practice · Medium

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