If $y = ({\log _{\cos x}}\sin x)({\log _{\sin x}}\cos x) + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}},$ then $\frac{{{\rm{d}}y}}{{{\rm{d}}x}}$ at $x = \frac{\pi }{2}$ is equal to

$\frac{8}{{\left( {4 + {\pi ^2}} \right)}}$

$ - \frac{8}{{\left( {4 + {\pi ^2}} \right)}}$

If $f(x) = {(|x|)^{|\sin \,x|}},$ then $f'( - \pi /4)$ is equal to

${\left( {\frac{\pi }{4}} \right)^{1/\sqrt 2 }}.\left( {\frac{{\sqrt 2 }}{4}\log \,\frac{4}{\pi } + \frac{{2\sqrt 2 }}{\pi }} \right)$

${\left( {\frac{\pi }{4}} \right)^{1/\sqrt 2 }}.\left( {\frac{{\sqrt 2 }}{2}\log \,\frac{\pi }{4} - \frac{{2\sqrt 2 }}{\pi }} \right)$

${\left( {\frac{\pi }{4}} \right)^{1/\sqrt 2 }}.\left( {\frac{{\sqrt 2 }}{2}\log \,\frac{\pi }{4} + \frac{{2\sqrt 2 }}{\pi }} \right)$

If $f\left( x \right) = 3{e^{{x^2}}}$, then $f'\left( x \right) - 2x{\rm{ }}f\left( x \right) + \frac{1}{3}f\left( 0 \right) - f'\left( 0 \right)$ is equal to

$\left( {7/3} \right){e^{{x^2}}}$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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Showing 151-180 of 402 questions

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151 Medium

151 of 402

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

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.

152 Medium

152 of 402

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.

153 Medium

153 of 402

Application +4 | -1

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.

154 Medium

154 of 402

Application +4 | -1

Let

y=y(x)

a

function of

x

satisfying

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

where

k

a

constant and

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

. Then

\frac{d y}{d x}

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.

155 Medium

155 of 402

Analyzing +4 | -1

f(x)=\cos x \cos 2 x \cos 2^2 x \cos 2^3 x \ldots \ldots

\cos 2^{n-1} x

and

n>1

, then

f^{\prime}\left(\frac{\pi}{2}\right)

, is

View answer and explanation

Correct option: A

Option D (None of these) is correct because

f'(x) = 0

for

x < 0

f'(-2) = 0

and

f'(-1)

is not defined.

156 Medium

156 of 402

Application +4 | -1

The derivative of

y=x^{\ln x}

View answer and explanation

Correct option: C

2 x^{\ln x-1} \ln x

Option C is correct because it is the result of logarithmic differentiation.

157 Medium

157 of 402

Analyzing +4 | -1

Let

\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}

be two functions defined by

f(x)=\left\{\begin{array}{ll} x \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0, & , x=0 \end{array}, \text { and } g(x)=x f(x)\right.

Statement I:

f

is a continuous function at

\mathrm{x}=0

.
Statement II:

g

is a differentiable function at

\mathrm{x}=0

View answer and explanation

Correct option: C

Statement

\mathrm{I}

is true, statement II is false.

Option C is correct because f(x) is continuous at x=0, but g(x) is not differentiable at x=0.

158 Medium

158 of 402

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)} \cdot f(\theta)

, is

View answer and explanation

Correct option: A

Option A (1) is correct because after simplification,

f(\theta) = \tan \theta

, and the derivative of

\tan \theta

with respect to

\tan \theta

is 1.

159 Medium

159 of 402

Analyzing +4 | -1

y = ({\log _{\cos x}}\sin x)({\log _{\sin x}}\cos x) + {\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}},

then

\frac{{\mathrm{d}y}}{{\mathrm{d}x}}

x = \frac{\pi }{2}

is equal to

View answer and explanation

Correct option: D

None of the above

Option D (None of the above) is correct because the function is not defined at

x = \frac{\pi}{2}

, so the derivative cannot be determined.

160 Medium

160 of 402

Application +4 | -1

The function

f(x)=e^x+x

, being differentiable and one-one, has a differentiable inverse

f^{-1}(x)

. The value of

\frac{d}{d x}\left(f^{-1}\right)

at the point

f(\log 2)

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because it is the value of the derivative of the inverse function at the specified point, calculated using the derivative of inverse functions.

161 Medium

161 of 402

Analyzing +4 | -1

x^{2}+y^{2}+\sin y=4

, then the value of

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

at the point

(-2,0)

View answer and explanation

Correct option: A

-34

Option A (-34) is correct because it is the value of the second derivative of y with respect to x, evaluated at the point (-2, 0).

162 Medium

162 of 402

Analyzing +4 | -1

f''\left( x \right) = - f\left( x \right)

, where

f\left( x \right)

is a continuous double differentiable function and

\mathrm{g}\left( x \right) = f'\left( x \right).\,

F\left( x \right) = {\left( {f\left( {\frac{x}{2}} \right)} \right)^2} + {\left( {\mathrm{g}\left( {\frac{x}{2}} \right)} \right)^2}

and

F\left( 5 \right) = 5

then

F\left( {10} \right)

View answer and explanation

Correct option: B

Option B (5) is correct because

F'(x) = 0

, indicating that

F(x)

is a constant function. Therefore,

F(10) = F(5) = 5

163 Medium

163 of 402

Application +4 | -1

u = \log \left( {\frac{{{x^2} + {y^2}}}{{x + y}}} \right),

then the value of

x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}

View answer and explanation

Correct option: C

Option C (1) is correct because the expression

x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}}

simplifies to 1 using Euler's theorem for homogeneous functions.

164 Medium

164 of 402

Application +4 | -1

The value of

\frac{d}{{dx}}\left( {\left| {x - 1} \right| + \left| {x - 5} \right|} \right)

x = 3

, is

View answer and explanation

Correct option: B

Option B (0) is correct because the derivative of the function at x=3 is 0, as the function is constant in the interval (1,5).

165 Medium

165 of 402

Application +4 | -1

f\left( x \right) = {e^x}\sin x,

then

f''\left( x \right)

is equal to

View answer and explanation

Correct option: B

2{e^x}\cos x

Option B (

2{e^x}\cos x

) is correct because it is the result of taking the second derivative of the given function

f(x) = e^x \sin x

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166 Medium

166 of 402

Application +4 | -1

f\left( x \right) = \sqrt {1 - \sin 2x}

, then

f'\left( x \right)

equals

View answer and explanation

Correct option: C

- \left( {\cos x + \sin x} \right),

for

x \in \left( {0,\pi /4} \right)

Option C (

- \left( {\cos x + \sin x} \right)

, for

x \in \left( {0,\pi /4} \right)

) is correct because it accounts for the absolute value and the interval in which the derivative is being evaluated.

167 Medium

167 of 402

Application +4 | -1

f(x)=x+\tan x

and

f

is inverse of

g

, then

g^{\prime}(x)

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because by differentiating the equation x = g(x) + tan(g(x)) and solving for g'(x), we obtain the given expression.

168 Medium

168 of 402

Application +4 | -1

The differential coefficient of the function

\left| {x - 1\left| + \right|x - 3} \right|

at the point

x = 2

View answer and explanation

Correct option: B

Option B (0) is correct because at x=2, the function f(x) is constant, hence its derivative is zero.

169 Medium

169 of 402

Application +4 | -1

f(x)=\log _x\{\ln (x)\}

, then

f^{\prime}(x)

x=e

View answer and explanation

Correct option: D

e^{-1}

Option D (

e^{-1}

) is correct because it is the value of the derivative of the function

f(x)

x = e

170 Medium

170 of 402

Application +4 | -1

The derivative of

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

with respect to

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

View answer and explanation

Correct option: B

Option B (1) is correct because both expressions simplify to

2 \tan^{-1}x

, making their derivatives equal, and the ratio of the derivatives is 1.

171 Medium

171 of 402

Analyzing +4 | -1

y(x)=x^{x}, x>0

, then

y^{\prime \prime}(2)-2 y^{\prime}

(2) is equal to:

View answer and explanation

Correct option: A

4\left(\log _{e} 2\right)^{2}-2

Option A is correct. By finding the first and second derivatives of

y(x) = x^x

and substituting

x=2

, we can evaluate

y''(2) - 2y'(2)

172 Medium

172 of 402

Application +4 | -1

x = \log \left( {1 + {t^2}} \right)

and

y = t - {\tan ^{ - 1}}t

.Then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because after finding the derivatives and substituting, the expression simplifies to

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

173 Medium

173 of 402

Application +4 | -1

g(x)=f\left(x-\sqrt{1-x^2}\right)

and

f^{\prime}(x)=1-x^2

then

g^{\prime}(x)

equals to :

View answer and explanation

Correct option: C

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

Option C (

2 x(x+\sqrt{1-x^2})

) is correct because it is the derivative of g(x) found by applying the chain rule and simplifying.

174 Medium

174 of 402

Application +4 | -1

y = \sqrt {x + \sqrt {x + \sqrt {x + ...\mathrm{to}\,\,\infty ,} } }

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{1}{{2y - 1}}

Option D (

\frac{1}{{2y - 1}}

) is correct because it is the derivative of the given function after simplifying and differentiating implicitly.

175 Medium

175 of 402

Application +4 | -1

y = {\tan ^{ - 1}}(\sec x - \tan x),

then

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

is equal to

View answer and explanation

Correct option: D

- \frac{1}{2}

Option D (-1/2) is correct because simplifying the expression inside the inverse tangent function and then differentiating gives -1/2.

176 Medium

176 of 402

Application +4 | -1

f(x)=\sqrt{\frac{1+\sin ^{-1} x}{1-\tan ^{-1} x}}

; then

f^{\prime}(0)

is equal to :

View answer and explanation

Correct option: D

Option D (1) is correct because the derivative at 0 is found by logarithmic differentiation and substituting x=0.

177 Medium

177 of 402

Application +4 | -1

x = \int \limits_0^y \frac{1}{{\sqrt {1 + 4{t^2}} }}dt

, then

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

View answer and explanation

Correct option: B

4y

Option B (

4y

) is correct because it is the result of differentiating the given integral equation twice and simplifying.

178 Medium

178 of 402

Application +4 | -1

f(x) = {(|x|)^{|\sin \,x|}},

then

f'( - \pi /4)

is equal to

View answer and explanation

Correct option: A

{\left( {\frac{\pi }{4}} \right)^{1/\sqrt 2 }}.\left( {\frac{{\sqrt 2 }}{2}\log \,\frac{4}{\pi } - \frac{{2\sqrt 2 }}{\pi }} \right)

Option A (

{\left( {\frac{\pi }{4}} \right)^{1/\sqrt 2 }}.{\left( {\frac{{\sqrt 2 }}{2}\log \frac{4}{\pi } - \frac{{2\sqrt 2 }}{\pi }} \right)}

) is correct because it is the value of the derivative of the function f(x) at x = -π/4, calculated using logarithmic differentiation.

179 Medium

179 of 402

Application +4 | -1

f\left( x \right) = 3{e^{{x^2}}}

, then

f'\left( x \right) - 2x\mathrm{ }f\left( x \right) + \frac{1}{3}f\left( 0 \right) - f'\left( 0 \right)

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because when the given expression is simplified using the derivative of the function, all terms cancel out except for a constant which evaluates to 1.

180 Medium

180 of 402

Analyzing +4 | -1

If for all

x,y \in R

, the function

f

is defined by

f\left( x \right) + f\left( y \right) + f\left( x \right)f\left( y \right) = 1

and

f\left( x \right) > 0

. Then,

View answer and explanation

Correct option: A

f'\left( x \right) = 0

for all

x \in R

Option A (

f'\left( x \right) = 0

for all

x \in R

) is correct because the given functional equation implies that

f(x)

is a constant function, hence its derivative is zero.

Revision notes and key takeaways

Review DIFFERENTIATION in your textbook or class notes, then use this set to check your understanding.

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