If $y = {\log _2}{\log _2}\left( x \right)$, then $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ is equal to

$\frac{{{{\log }_2}e}}{{{{\log }_e}x}}$

$\frac{{{{\log }_2}e}}{{x{{\log }_x}2}}$

$\frac{{{{\log }_2}x}}{{{{\log }_e}2}}$

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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 61-90 of 402 questions

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61 Medium

61 of 402

Application +4 | -1

f'(x) = g(x)

and

g'(x) = - f(x)

for all real

x

and

f(5) = 2 = f'(5),

then

{f^2}(10) + {g^2}(10)

equals

View answer and explanation

Correct option: C

Option C (8) is correct because the expression

{f^2}(10) + {g^2}(10)

is constant and equal to

{f^2}(5) + {g^2}(5)

, which evaluates to 8.

62 Medium

62 of 402

Application +4 | -1

Let

f(x)=e^x, g(x)=e^{-x}

and

h(x)=g(f(x))

. Then

h^{\prime}(0)

is equal to

View answer and explanation

Correct option: A

\frac{-1}{e}

Option A (-1/e) is correct because it is the value of the derivative of h(x) = e^{(-e^x)} evaluated at x=0.

63 Medium

63 of 402

Analyzing +4 | -1

y=f(x)

and

x=g(y)

are inverse functions of each other, then

View answer and explanation

Correct option: B

g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^3}

Option B (

g^{\prime \prime}(y)=-\frac{f^{\prime \prime}(x)}{\left(f^{\prime}(x)\right)^3}

) is correct because it is the correct second derivative of the inverse function, derived using the chain rule.

64 Medium

64 of 402

Application +4 | -1

\sin (x+y)=\log (x+y)

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: D

-1

Option D is correct because the derivative of the implicit function simplifies to -1.

65 Medium

65 of 402

Application +4 | -1

y = {\cos ^{ - 1}}\left( {\frac{{1 - \log x}}{{1 + \log x}}} \right),

then

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

x = e

View answer and explanation

Correct option: C

\frac{1}{{2e}}

Option C (

\frac{1}{{2e}}

) is correct because after differentiating and substituting

x = e

, the derivative evaluates to

\frac{1}{{2e}}

66 Medium

66 of 402

Application +4 | -1

Let

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

for all real

x

and

y

for

f^{\prime}(0)=-1, f(0)=1

, then

f^{\prime}(2)=

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because the given functional equation implies that

f'(x)

is constant, and since

f'(0) = -1

, then

f'(2) = -1

67 Medium

67 of 402

Application +4 | -1

Differential co-efficient of

\sin ^{-1} x

w.r.t

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

View answer and explanation

Correct option: A

Option A (1) is correct because the derivative of

\sin^{-1}x

with respect to

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

is 1, as they are essentially the same function.

68 Medium

68 of 402

Application +4 | -1

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

then

y'\left( 0 \right)

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

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

x=0

after simplifying using trigonometric substitution.

69 Medium

69 of 402

Application +4 | -1

y = {\log _2}{\log _2}\left( x \right)

, then

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

is equal to

View answer and explanation

Correct option: D

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

The correct answer is D (

\frac{{{{\log }_2}e}}{{x{{\log }_e}x}}

) because it is the result of correctly applying the chain rule and change of base formula for logarithms.

70 Medium

70 of 402

Application +4 | -1

f\left( x \right) = \left( {x - 2} \right)\left( {x - 4} \right)\left( {x - 6} \right) \ldots \left( {x - 2n} \right),

then

f'\left( 2 \right)

View answer and explanation

Correct option: B

{\left( { - 2} \right)^{n - 1}}\left( {n - 1} \right)!

Option B (

{(- 2)^{n - 1}}(n - 1)!

) is correct because after differentiating the given function and evaluating at x=2, the result is

{(- 2)^{n - 1}}(n - 1)!

71 Medium

71 of 402

Application +4 | -1

f(x)=\sqrt{1+\cos ^{2}\left(x^{2}\right)}

, then

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

View answer and explanation

Correct option: B

-\sqrt{(\pi / 6)}

The correct answer is B (

-\sqrt{(\pi / 6)}

) because it is the value of the derivative of the given function evaluated at

x=\frac{\sqrt{\pi}}{2}

, using the chain rule and the derivatives of trigonometric functions.

72 Medium

72 of 402

Application +4 | -1

f:(0, \infty) \rightarrow \mathbb{R}

is continuous. If

F(x)

is a differentiable function such that

F^{\prime}(x)=f(x) \forall x>0

and

F\left(x^2\right)=x^2+x^3

, then

f(4)

equals

View answer and explanation

Correct option: C

Option C (4) is correct because by differentiating F(x²) and using F'(x) = f(x), we can solve for f(4).

73 Medium

73 of 402

Application +4 | -1

\sin y + {e^{ - x\cos y}} = e

, then

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

\left( {1,\pi } \right)

View answer and explanation

Correct option: C

e

Option D (

e

) is correct because after differentiating the given equation implicitly and substituting x=1 and y=

\pi

, the value of

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

is found to be e.

74 Medium

74 of 402

Analyzing +4 | -1

Let

y = {t^{10}} + 1

and

x = {t^8} + 1,

then

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

is equal to

View answer and explanation

Correct option: C

\frac{5}{{16{t^6}}}

Option C (

\frac{5}{{16{t^6}}}

) is correct because after finding the first and second derivatives using parametric differentiation, the expression simplifies to this result.

75 Medium

75 of 402

Application +4 | -1

{10^{ - x\tan x}}\left[ {\frac{d}{{dx}}\left( {{{10}^{x\tan x}}} \right)} \right]

is equal to

View answer and explanation

Correct option: B

In 10

(\tan x + x{\sec ^2}x)

Option B (In 10

(\tan x + x{\sec ^2}x)

) is correct because it is the result of differentiating

10^{x\tan x}

and then simplifying.

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76 Medium

76 of 402

Analyzing +4 | -1

Let

y(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)\left(1+x^{16}\right)

. Then

y^{\prime}-y^{\prime \prime}

x=-1

is equal to :

View answer and explanation

Correct option: A

496

Option A (496) is correct because after simplifying the expression for y(x), we can find the first and second derivatives and evaluate them at x = -1, then compute y' - y''.

77 Medium

77 of 402

Application +4 | -1

f(x)=\frac{x^2-x}{x^2+2 x}

, then

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

is equal to

View answer and explanation

Correct option: B

\frac{3}{(1-x)^2}

Option B (

\frac{3}{(1-x)^2}

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

f^{-1}(x)

after simplifying

f(x)

and finding its inverse.

78 Medium

78 of 402

Understanding +4 | -1

f(x)=\left(\frac{x^a}{x^b}\right)^{a+b} \cdot\left(\frac{x^b}{x^c}\right)^{b+c} \cdot\left(\frac{x^c}{x^a}\right)^{c+a}

then

f^{\prime}(x)

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because after simplification, f(x) becomes 1, and the derivative of a constant is 0.

79 Medium

79 of 402

Analyzing +4 | -1

Let

f:R \to R

be a function such that

f(x) = {x^3} + {x^2}f'(1) + xf''(2) + f'''(3),x \in R

Then

f(2)

equals

View answer and explanation

Correct option: C

-2

Option C (-2) is correct because by finding the derivatives and solving the resulting system of equations, we can determine f(x) and then evaluate f(2).

80 Medium

80 of 402

Application +4 | -1

f(x)

and

g(x)

are two functions from

\mathbb{R}

\mathbb{R}

such that

(f \circ g)(x)=\left(x^3-x^2+2\right)^8

, then

f^{\prime}(1) g^{\prime}(1)

View answer and explanation

Correct option: A

Option A (8) is correct because, by choosing

f(x) = x^8

and

g(x) = x^3 - x^2 + 2

, we get

(f \circ g)(x) = (x^3 - x^2 + 2)^8

. Then,

f'(1) = 8

and

g'(1) = 1

, so

f'(1)g'(1) = 8

81 Medium

81 of 402

Application +4 | -1

Let

f(x)=\frac{x^{2}-x}{x^{2}+2 x} x \neq 0,-2

. Then

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

(wherever it is defined) is equal to :

View answer and explanation

Correct option: B

\frac{3}{(1-x)^{2}}

Option B (

\frac{3}{(1-x)^{2}}

) is correct because it is the derivative of the inverse function after simplification.

82 Medium

82 of 402

Application +4 | -1

f(x)=\sqrt{1+\cos ^2 x^2}

, then

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

View answer and explanation

Correct option: B

-\sqrt{\frac{\pi}{6}}

Option B (

-\sqrt{\frac{\pi}{6}}

) is correct because it is the result of differentiating the given function and substituting

x = \frac{\sqrt{\pi}}{2}

83 Medium

83 of 402

Application +4 | -1

\sqrt{x+y}+\sqrt{y-x}=c

, then

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

is equal to

View answer and explanation

Correct option: C

-\frac{2}{c^2}

Option C (

\frac{2}{c^2}

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

84 Medium

84 of 402

Application +4 | -1

y = {\sin ^{ - 1}}\left( {\frac{{\sin \alpha \sin x}}{{1 - \cos \alpha \sin x}}} \right)

, then

y'\left( 0 \right)

View answer and explanation

Correct option: D

\sin \alpha

Option D (

\sin \alpha

) is the correct derivative at x=0, obtained by differentiating the given function and substituting x=0.

85 Medium

85 of 402

Application +4 | -1

{e^{y + {e^{y + \ldots \infty }}}},x > 0

then

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

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the derivative obtained by implicit differentiation of the given equation.

86 Medium

86 of 402

Application +4 | -1

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

, then

\left(1-\mathrm{x}^2\right) \mathrm{y_2-x y_1}=

View answer and explanation

Correct option: B

Option B (4) is correct because, after differentiating the given equation twice and simplifying, the expression (1-x²)y_2 - xy_1 equals 4.

87 Medium

87 of 402

Application +4 | -1

x y=a x^2+(b / x)

, then

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

View answer and explanation

Correct option: C

\frac{2 y}{x}

Option C (

\frac{{2 y}}{{x}}

) is correct because after differentiating the given equation twice and simplifying, the expression

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

simplifies to

\frac{{2 y}}{{x}}

88 Medium

88 of 402

Analyzing +4 | -1

Let S be the set of all points in

(-\pi, \pi)

at which the function

f(x)=\min |\sin x, \cos x|

is not differentiable. Then

S

is a subset of which of the following?

View answer and explanation

Correct option: A

\left\{\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}\right\}

Option A is correct because the function is not differentiable at points where the absolute values of sine and cosine intersect within the given interval, which are

\frac{-3 \pi}{4}, \frac{-\pi}{4}, \frac{\pi}{4}, \frac{3 \pi}{4}

89 Medium

89 of 402

Understanding +4 | -1

Let

g\left( x \right)

be the inverse of an invertible function

f\left( x \right)

which is differentiable at

x = c

, then

g'\left( {f\left( c \right)} \right)

equals

View answer and explanation

Correct option: B

\frac{1}{{f'\left( c \right)}}

Option B (

\frac{1}{{f'(c)}}

) is correct because it is the direct application of the inverse function derivative rule.

90 Medium

90 of 402

Application +4 | -1

x^y \cdot y^x=16

, then

\frac{d y}{d x}

(2,2)

View answer and explanation

Correct option: A

-1

Option A (-1) is correct because it is the value of

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

at the point (2,2) after differentiating the given equation implicitly.

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