$y y^{\prime \prime}-2\left(y^{\prime}\right)^2+1=0$

$y y^{\prime \prime}+2\left(y^{\prime}\right)^2+1=0$

If $y = {\log _{\cos x}}\sin x,$ then $\frac{{dy}}{{dx}}$ is equal to

$\frac{{(\tan x\log \cos x + \cot x\log \sin x)}}{{{{(\log \cos x)}^2}}}$

$\frac{{(\cot x\,{\rm{log cos}}\,x + \tan x\log \sin x)}}{{{{(\log \sin x)}^2}}}$

If $f\left( x \right) = x\left( {\sqrt x + \sqrt {x + 1} } \right)$, then

$f\left( x \right)$ is continuous but not differentiable at $x = 0$

$f\left( x \right)$ is differentiable at $x = 0$

If $y=(1+x)^{1 / x}$, then $\frac{d y}{d x}=$

$\frac{x}{1+x}(1+x)^{1 / x}-\log (1+x)$

The function $f$ defined by $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{\sin {x^2}}}{x},x \ne 0}\\ {0,x = 0} \end{array}} \right.$ is

Neither continuous nor derivable at $x = 0$

Continuous but not derivable at $x = 0$

On maximum point of ${\sin ^p}x\,{\cos ^q}x$ is

$x = {\tan ^{ - 1}}\left( {\sqrt {\frac{q}{p}} } \right)$

$x = {\tan ^{ - 1}}\left( {\frac{p}{q}} \right)$

$x = {\tan ^{ - 1}}\left( {\frac{q}{p}} \right)$

If $y = {\cos ^{ - 1}}(\cos x),$ then $\frac{{dy}}{{dx}}$

${\rm{ - 1}}$ in the whole plane

1 in the 2nd and 3rd quadrants of the plane

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 91-120 of 402 questions

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91 Medium

91 of 402

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because it is the derivative of the simplified form of the given function after using trigonometric substitution.

92 Medium

92 of 402

Application +4 | -1

x^2+y^2=1

, then

View answer and explanation

Correct option: B

y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0

Option B is correct because it is the result of differentiating the given implicit equation twice and rearranging the terms.

93 Medium

93 of 402

Understanding +4 | -1

f(x)=(1+x)^{n }

, then

f(0)+f(0)+\frac{1}{2} f^{\prime}(0)+\ldots+\frac{1}{n !} f^n(0)=\ldots \ldots

View answer and explanation

Correct option: B

2^n

Option B (

2^n

) is correct because the given expression is the binomial expansion of

(1+1)^n

94 Medium

94 of 402

Application +4 | -1

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

, then

\frac{d y}{d x}

View answer and explanation

Correct option: B

Option B is correct because it simplifies to 1 after applying trigonometric identities and inverse cosine properties.

95 Medium

95 of 402

Application +4 | -1

y = {\log _{\cos x}}\sin x,

then

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

is equal to

View answer and explanation

Correct option: A

\frac{{(\cot x\log \cos x + \tan x\log \sin x)}}{{{{(\log \cos x)}^2}}}

Option A is correct because it correctly applies the quotient rule and change of base formula for logarithms.

96 Medium

96 of 402

Analyzing +4 | -1

f\left( x \right) = x\left( {\sqrt x + \sqrt {x + 1} } \right)

, then

View answer and explanation

Correct option: C

f\left( x \right)

is not differentiable at

x = 0

Option C is correct because the function is not continuous at x=0 due to the square root of x, and therefore not differentiable.

97 Medium

97 of 402

Application +4 | -1

x = \exp \left\{ {{{\tan }^{ - 1}}\left( {\frac{{y - {x^2}}}{{{x^2}}}} \right)} \right\}

, then

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

equals

View answer and explanation

Correct option: A

2x[1 + \tan (\log x)] + x{\sec ^2}(\log x)

Option A (

2x[1 + \tan (\log x)] + x{\sec ^2}(\log x)

) is correct because it is the result of differentiating the given equation and solving for dy/dx.

98 Medium

98 of 402

Application +4 | -1

y=(1+x)^{1 / x}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: B

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

Option B (

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

) is correct because after applying logarithmic differentiation and simplifying, the derivative is found to be

(1+x)^{1 / x} \frac{\left[\frac{x}{1+x}-\log (1+x)\right]}{x^2}

99 Medium

99 of 402

Application +4 | -1

Let

f(x)=2 /(x+1)

and

g(x)=3 x

. It is given that

(f \circ g)\left(x_0\right)=(g \circ f)\left(x_0\right)

. Then

(g \circ f)^{\prime}\left(x_0\right)

equals

View answer and explanation

Correct option: D

\frac{-32}{3}

Option D (

\frac{-32}{3}

) is correct because, after finding

x_0 = -\frac{1}{4}

, we use the chain rule to compute

(g \circ f)'(x_0) = g'(f(x_0)) * f'(x_0)

, which results in

\frac{-32}{3}

100 Medium

100 of 402

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: B

\tan u

Option B (

\tan u

) is correct because Euler's theorem is applied to the homogeneous function

f(x,y) = \sin u

to find the value of

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

101 Medium

101 of 402

Application +4 | -1

f(x)=\cos \left(x^2-4[x]\right)

for

0 < x < 1

, where

[x]=

G.I.F

., then

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

is equal to

View answer and explanation

Correct option: A

-\sqrt{\frac{\pi}{2}}

Option A (

-\sqrt{\frac{{\pi}}{{2}}}

) is correct because it is the value of the derivative of the function at the given point.

102 Medium

102 of 402

Analyzing +4 | -1

Let

f

and

g

be twice differentiable functions on

\mathrm{R}

such that

f^{\prime \prime}(x)=g^{\prime \prime}(x)+6 x

f^{\prime}(1)=4 g^{\prime}(1)-3=9

f(2)=3 g(2)=12 .

Then which of the following is NOT true?

View answer and explanation

Correct option: D

-1 < x < 2

, then

|f(x)-g(x)| < 8

Option D is NOT true because when x = -1, |f(x) - g(x)| = 10, which is greater than 8, invalidating the statement that |f(x) - g(x)| < 8 for -1 < x < 2.

103 Medium

103 of 402

Analyzing +4 | -1

Let

g(x)=\log f(x)

, where

f(x)

is twice differentiable positive function on

(0, \infty)

such that

f(x+1)=x f(x)

. Then, for

N=1,2,3, \ldots g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=

View answer and explanation

Correct option: A

-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}

Option A (

-4\left\{1+\frac{1}{9}+ \frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}

) is correct because it represents the difference in the second derivative of

g(x)

evaluated at

N+\frac{1}{2}

and

\frac{1}{2}

, derived using the given functional equation.

104 Medium

104 of 402

Analyzing +4 | -1

The function

f

defined by

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{\sin {x^2}}}{x},x \ne 0}\\ {0,x = 0} \end{array}} \right.

View answer and explanation

Correct option: A

Continuous and derivable at

x = 0

Option A (Continuous and derivable at

x = 0

) is correct because the function is both continuous and differentiable at the specified point.

105 Medium

105 of 402

Application +4 | -1

\mathrm{x}=\frac{\mathrm{t}-\mathrm{e}^{-\mathrm{t}^2}}{2 \mathrm{t}^2}

, then

\left(\mathrm{t} \frac{\mathrm{dx}}{\mathrm{dt}}-\mathrm{e}^{-\mathrm{t}^2}\right)=

View answer and explanation

Correct option: A

\frac{1}{2 \mathrm{t}}-2 \mathrm{x}

Option A (

\frac{1}{2 \mathrm{t}}-2 \mathrm{x}

) is correct because it's the result of correctly differentiating x with respect to t and then substituting back the original expression for x.

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106 Medium

106 of 402

Analyzing +4 | -1

On maximum point of

{\sin ^p}x\,{\cos ^q}x

View answer and explanation

Correct option: A

x = {\tan ^{ - 1}}\left( {\sqrt {\frac{p}{q}} } \right)

Option A is correct because the maximum point is found by setting the derivative of the function to zero and solving for x.

107 Medium

107 of 402

Understanding +4 | -1

f:R \to R

is an even function which is twice differentiable on

R

and

f''\left( \pi \right) = 1,

then

f''( - \pi )

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because the second derivative of an even function is also an even function, and therefore

f''(-\pi) = f''(\pi) = 1

108 Medium

108 of 402

Application +4 | -1

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

for all

x,y,z

and

f\left( 2 \right) = 4,f'\left( 0 \right) = 3

, then

f'\left( 2 \right)

equals

View answer and explanation

Correct option: A

Option A (12) is correct because f'(2) = 4f'(0) = 4 * 3 = 12.

109 Medium

109 of 402

Analyzing +4 | -1

Let

f(x)=

Degree of

\left(u^{x^2}+u^2+2 u+3\right)

. Then, at

x=\sqrt{2}, f(x)

View answer and explanation

Correct option: A

continuous but not differentiable

Option A (continuous but not differentiable) is correct because the function is continuous at x=

\sqrt{2}

but the left-hand and right-hand derivatives are not equal at that point.

110 Medium

110 of 402

Application +4 | -1

The value of the derivative of

|x-1|+|x-3|

x=2

View answer and explanation

Correct option: B

Option B (0) is correct because at x=2, the function f(x) = |x-1| + |x-3| is differentiable, and its derivative is 0.

111 Medium

111 of 402

Application +4 | -1

f(x)=x^3+x^4+\log x

and

g

is the inverse of

f

, then

g^{\prime}(2)

is :

View answer and explanation

Correct option: B

\frac{1}{8}

Option B (

\frac{1}{8}

) is correct because it is the value of

g'(2)

found by using the inverse function derivative formula and evaluating at the correct point.

112 Medium

112 of 402

Application +4 | -1

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

, then

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

is equal to

View answer and explanation

Correct option: B

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

Option B is correct because it correctly applies the quotient rule and chain rule to differentiate the logarithmic function with a variable base.

113 Medium

113 of 402

Application +4 | -1

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

equals

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the correct application of the chain rule and inverse function differentiation.

114 Medium

114 of 402

Analyzing +4 | -1

y = \sqrt {x + \sqrt {y + \sqrt {x + \sqrt {y + \ldots \infty } } } }

, then

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

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated derivative

\frac{{dy}}{{dx}} = \frac{{{{y^2} - x}}}{{{{y^3} - xy - 1}}}

does not match any of the given options.

115 Medium

115 of 402

Application +4 | -1

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

then

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

is equal to

View answer and explanation

Correct option: D

3\tan u

Option D (

3\tan u

) is correct because, by applying Euler's theorem for homogeneous functions, the expression

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

simplifies to

3\tan u

116 Medium

116 of 402

Analyzing +4 | -1

Let

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

and,

g(x)=x^2+x f^{\prime}(2)+f^{\prime \prime}(3)

. Then,

View answer and explanation

Correct option: D

All the above

Option D (All the above) is correct because all the given relationships between the derivatives of f(x) and g(x) are satisfied.

117 Medium

117 of 402

Analyzing +4 | -1

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

then

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

View answer and explanation

Correct option: D

\mathrm{ - 1}

in the 3rd and 4th quadrants of the plane

The correct answer is D (

\mathrm{ - 1}

in the 3rd and 4th quadrants of the plane) because the derivative of

y = \cos^{-1}(\cos x)

is -1 for

\pi < x \le 2\pi

, which corresponds to the 3rd and 4th quadrants.

118 Medium

118 of 402

Analyzing +4 | -1

Let

f: R \rightarrow R

be a function such that

f\left( {\frac{{x + y}}{3}} \right) = \frac{{f(x) + f(y)}}{3}\,\,\mathrm{ for}\,\,\mathrm{ all }\,\,x,y \in R

f(0)=0

and

f^{\prime}(0)=3

. Then,

f(x)

View answer and explanation

Correct option: C

differentiable in

R

The correct answer is C (differentiable in

R

) because the given functional equation and differentiability at 0 imply that

f(x)

is a linear function, hence differentiable everywhere.

119 Medium

119 of 402

Application +4 | -1

Let

f(x)=\frac{\sin x+\cos x -\sqrt{2}}{\sin x-\cos x}, x \in[0, \pi]-\left\{\frac{\pi}{4}\right\}

. Then

f\left(\frac{7 \pi}{12}\right) f^{\prime \prime}\left(\frac{7 \pi}{12}\right)

is equal to

View answer and explanation

Correct option: A

\frac{2}{9}

Option A (2/9) is correct because it is the result of correctly simplifying the function and calculating the required values and their product.

120 Medium

120 of 402

Analyzing +4 | -1

A continuous and differentiable function

y=f(x)

is such that it's graph cuts line

y=m x+c

n

distinct points. Then the minimum number of points at which

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

View answer and explanation

Correct option: C

{n}-2

Option C (

{n}-2

) is correct because by applying LMVT and Rolle's theorem, we can deduce that there are at least n-2 points where the second 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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