If $ y=\sqrt{\frac{1+\tan x}{1-\tan x}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \sec \left(\frac{\pi}{4}+x\right) $

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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 181-210 of 402 questions

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181 Medium

181 of 402

Analyzing +4 | -1

Let

g(x)=\log _e f(x)

, where

f(x)

is twice differentiable positive function on

(0, \infty)

such that

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

, for

N=1,2,3, \ldots

, then

g''\left( {N + \frac{1}{2}} \right) - g''\left( {\frac{1}{2}} \right)\mathrm{ equals }

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 is the result obtained after differentiating the given logarithmic function and summing the series.

182 Medium

182 of 402

Analyzing +4 | -1

The value of

\frac{d}{{dx}}\left[ {\left( {\frac{{{{\tan }^2}2x - {{\tan }^2}x}}{{1 - {{\tan }^2}2x{{\tan }^2}x}}} \right)\cot 3x} \right]

View answer and explanation

Correct option: C

{\sec ^2}x

The correct answer is C (

{\sec ^2}x

) because the expression simplifies to

\tan x

after applying trigonometric identities, and the derivative of

\tan x

{\sec ^2}x

183 Medium

183 of 402

Application +4 | -1

\ln \left(({e}-1){e}^{{xy}}+{x}^2\right)={x}^2+{y}^2

, then

\left(\frac{{dy}}{{dx}}\right)_{(1,0)}

is given by

View answer and explanation

Correct option: A

Option A (2) is correct because by differentiating the given equation implicitly and substituting the point (1,0), we can find the value of

\frac{dy}{dx}

184 Medium

184 of 402

Application +4 | -1

Differential coefficient of

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

with respect to

\sqrt {1 - {x^2}}

x = \frac{1}{2}

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because after substituting

x = \frac{1}{2}

, the ratio of the derivatives simplifies to 4.

185 Medium

185 of 402

Analyzing +4 | -1

{x^2} + {y^2} = t - \frac{1}{t}

and

{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}}

, then

{x^3}y\frac{{dy}}{{dx}}

equals

View answer and explanation

Correct option: B

The correct answer is 1 because after simplifying the given equations and differentiating,

{x^3}y\frac{{dy}}{{dx}}

equals 1.

186 Medium

186 of 402

Application +4 | -1

y = \sqrt {\log x + \sqrt {\log x + \sqrt {\log x + \sqrt {\log x + ...\infty } } } ,}

then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

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

187 Medium

187 of 402

Application +4 | -1

y=\sqrt{\frac{1+\tan x}{1-\tan x}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \sec ^{2}\left(\frac{\pi}{4}+x\right)

Option A is correct because the derivative is found using the chain rule and trigonometric identities to simplify the expression.

188 Medium

188 of 402

Application +4 | -1

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

, then the value of

f'\left( {\frac{{\sqrt \mathrm{\pi } }}{2}} \right)

View answer and explanation

Correct option: B

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

Option B (

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

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

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

189 Medium

189 of 402

Application +4 | -1

A continuous and differentiable function

f(x)

satisfy the functional equation

f(x+y)=\frac{f(x) f(y)}{f(x)+f(y)}

f(1)=32

then

f(4)=

View answer and explanation

Correct option: A

Option A (8) is correct because by substituting the functional equation into a simpler form and using the given condition

f(1)=32

, we can find the value of

f(4)

190 Medium

190 of 402

Analyzing +4 | -1

{x^2} + {y^2} = t - \frac{1}{t}

and

{x^4} + {y^4} = {t^2} + \frac{1}{{{t^2}}},

then

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

is equal to

View answer and explanation

Correct option: D

\frac{1}{{{x^3}y}}

The correct answer is D (

\frac{{1}}{{{{x^3}y}}}

) because it is obtained by differentiating the given equations and solving for

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

191 Medium

191 of 402

Application +4 | -1

The derivative of

f(x)=\log _{x^2} 3

, w. r. to

x

is.

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of applying the chain rule and change of base formula to differentiate the given logarithmic function.

192 Medium

192 of 402

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 is correct because it correctly applies the chain rule and the derivative of an inverse function.

193 Medium

193 of 402

Application +4 | -1

The derivative of

{\sin ^2}x

with respect to

{\cos ^2}x

View answer and explanation

Correct option: D

None of these

The correct answer is None of these because the derivative of sin²x with respect to cos²x is -1, which is not among the given options.

194 Medium

194 of 402

Analyzing +4 | -1

Let

f(x)=-4 \cdot e^{\frac{2-x}{4}}+10+2 x+x^3+x^5

and

g(x)=f^{-1}(x)

, then the value of

g^{\prime}(50)

View answer and explanation

Correct option: C

\frac{1}{95}

Option C (

\frac{1}{95}

) is correct because using the inverse function derivative formula

g'(x) = \frac{1}{f'(g(x))}

, we find

g'(50) = \frac{1}{f'(2)} = \frac{1}{95}

195 Medium

195 of 402

Analyzing +4 | -1

Let

\left( x \right) = \log f\left( x \right),

where

f\left( x \right)

is a twice differentiable positive function on

\left( {0,\infty } \right)

such that

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

Then, for

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

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the result obtained after substituting and simplifying the given expression.

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

196 Medium

196 of 402

Application +4 | -1

\frac{d}{d x}\left(\frac{\sec x+\tan x}{\sec x-\tan x}\right)

is equal to

View answer and explanation

Correct option: A

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

Option A is correct because the expression is simplified using trigonometric identities and then differentiated.

197 Medium

197 of 402

Analyzing +4 | -1

Let

g\left( x \right) = \frac{{{{\left( {x - 1} \right)}^n}}}{{\log {{\cos }^m}\left( {x - 1} \right)}},0 < x < 2,m

and

n

are integers,

m \ne 0,n > 0

, and let

p

be the left hand derivative of

\left| {x - 1} \right|

x = 1

. If

\lim\limits_{x \to {1^ + }} g\left( x \right) = p

, then

View answer and explanation

Correct option: C

n = 2,m = 2

Option C (

n = 2,m = 2

) is correct because it satisfies the limit condition

\lim\limits_{{x \to {1^ + }}} g(x) = p = -1

198 Medium

198 of 402

Application +4 | -1

2 y=\left(\cot ^{-1}\left(\frac{\sqrt{3} \cos x+\sin x}{\cos x-\sqrt{3} \sin x}\right)\right)^{2}, x \in\left(0, \frac{\pi}{2}\right)

then

\frac{d y}{d x}

is equal to :

View answer and explanation

Correct option: A

\frac{\pi}{6}-x

Option A (

\frac{\pi}{6}-x

) is correct because by simplifying the expression inside the cotangent inverse function, we get

2y = (\frac{\pi}{6} - x)^2

, and differentiating this gives

\frac{dy}{dx} = \frac{\pi}{6}-x

199 Medium

199 of 402

Analyzing +4 | -1

f\left( x \right) = ({\log _{\cot x}}{\log _{\cot x}}\tan x)({\log _{\tan x}}{\log _{\tan x}}\cot {\left. x \right)^{ - 1}} + {\tan ^{ - 1}}\frac{{4x}}{{4 - {x^2}}},

then

f'\left( 2 \right)

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}

Option A (

\frac{1}{2}

) is correct because after simplifying the function and differentiating, the value of the derivative at x=2 is

\frac{1}{2}

200 Medium

200 of 402

Application +4 | -1

f\left( x \right) = \left( {x + 1} \right){\tan ^{ - 1}}\left( {{e^{ - 2x}}} \right)

, then

f'\left( 0 \right)

View answer and explanation

Correct option: B

\frac{\pi }{4} - 1

Option B (

\frac{\pi }{4} - 1

) is correct because after differentiating and substituting

x = 0

, the value of

f'(0)

\frac{\pi }{4} - 1

201 Medium

201 of 402

Analyzing +4 | -1

Let

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

and

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

for all

x,y \in R

, where

g\left( x \right)

is continuous function. Then,

f'\left( x \right)

is equal to

View answer and explanation

Correct option: D

Option D (0) is correct because, given f(x+y) = f(x) + f(y) and f(x) = x²g(x), it implies f(x) = 0 for all x, thus f'(x) = 0.

202 Medium

202 of 402

Analyzing +4 | -1

F\left( x \right) = \frac{1}{{{x^2}}}\int \limits_4^x \left( {4{t^2} - 2F'\left( t \right)} \right)dt

, then

F'\left( 4 \right)

equals

View answer and explanation

Correct option: A

\frac{{32}}{9}

Option A (

\frac{{32}}{9}

) is correct because after differentiating both sides and substituting x=4, we find that F'(4) =

\frac{{32}}{9}

203 Medium

203 of 402

Analyzing +4 | -1

x=2 \sin \theta-\sin 2 \theta

and

y=2 \cos \theta-\cos 2 \theta, \theta \in[0,2 \pi]

, then

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

\theta=\pi

is :

View answer and explanation

Correct option: B

\frac{3}{8}

Option B (

\frac{3}{8}

) is correct because it is the value of the second derivative of y with respect to x, evaluated at

\theta = \pi

204 Medium

204 of 402

Application +4 | -1

y=\sin x \cdot \sin 2 x \cdot \sin 3 x \ldots \cdot \sin n x

, then

\frac{d y}{d x}=\ldots \ldots

View answer and explanation

Correct option: B

y \sum\limits_{{k=1}}^n k \cot k x

Option B (

y \sum\limits_ {{k=1}}^n k \cot k x

) is correct because by taking the logarithm of both sides and differentiating, we can express dy/dx in terms of y and a sum of cotangents.

205 Medium

205 of 402

Application +4 | -1

\frac{d}{{dx}}\left[ {\log \left\{ {{e^x}{{\left( {\frac{{x - 2}}{{x + 2}}} \right)}^{\frac{3}{4}}}} \right\}} \right]

is equal to

View answer and explanation

Correct option: C

\frac{{{x^2} - 1}}{{{x^2} - 4}}

Option C (

\frac{{{x^2} - 1}}{{{x^2} - 4}}

) is correct because it is the result of applying logarithmic and derivative rules to the given expression.

206 Medium

206 of 402

Analyzing +4 | -1

f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1\end{array}\right|

, then

\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}

View answer and explanation

Correct option: A

Exists and is equal to -2 [Online April 15, 2018]

Option A (Exists and is equal to -2) is correct because the limit of f'(x)/x as x approaches 0 is -2, obtained by applying L'Hopital's rule or by simplifying the expression for f'(x).

207 Medium

207 of 402

Analyzing +4 | -1

f:(0, \infty) \rightarrow(0, \infty)

is a twice differentiable function and satisfies the functional relation

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

for all positive

x

. If

g(x)=\log (f(x))

, then for any positive integer

n

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}+\cdots+\frac{1}{(2 n-1)^2}\right)

Option A is correct because by differentiating the given functional relation twice and substituting

x = 1, 2, ..., n

, the expression simplifies to

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

208 Medium

208 of 402

Application +4 | -1

y = {\log ^n}x

, where

{\log ^n}

means

\log \log \log \ldots

(repeated

n

times ),then

x\log x{\log ^2}x{\log ^3}x \ldots {\log ^{n - 1}}x{\log ^n}x\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: D

{\log ^n}x

Option D (

{\log ^n}x

) is correct because after differentiating and rearranging, the expression simplifies to

{\log ^n}x

209 Medium

209 of 402

Application +4 | -1

f(x)=\sqrt{x^2+9}

, then

\lim _{x \rightarrow 4} \frac{f(x)-f(4)}{x-4}

has the value

View answer and explanation

Correct option: C

4 / 5

Option C (

4 / 5

) is correct because the limit is the definition of the derivative of

f(x)

x=4

, and evaluating

f'(4)

gives

4/5

210 Medium

210 of 402

Application +4 | -1

Let

f(x)

be a polynomial of second degree. If

f(1)=f(-1)

and

a, b

, care in A.P., then

f^{\prime}(a), f^{\prime}(b), f^{\prime}(c)

are in

View answer and explanation

Correct option: B

Option B is correct because it correctly uses the given conditions to determine that the derivatives form an arithmetic progression.

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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