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

If $y=x+e^x$, then $\frac{d^2 x}{d y^2}$ is equal to

$\frac{1}{\left(1+e^x\right)^2}$

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

Let $f: R \rightarrow R$ be a function such that $f\left( {\frac{{x + y}}{3}} \right) = \frac{{f(x) + f(y)}}{3}\,\,{\rm{ for}}\,\,{\rm{ all }}\,\,x,y \in R$ $f(0)=0$ and $f^{\prime}(0)=3$. Then, $f(x)$ is

continuous but not differentiable

If $f(x)=\left|\log _e\right| x||$ then $f^{\prime}(x)$ equals

$\frac{1}{x}$ for $x>0$ and $-\frac{1}{x}$ for $x<0$

The derivative of $\log \left| x \right|$ is

$\frac{1}{{\left| x \right|}},x \ne 0$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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Showing 241-270 of 932 questions

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241 Medium

241 of 932

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

242 Easy

242 of 932

Application +4 | -1

for

0 < x < 2,\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}\sqrt {\frac{{1 + \cos \frac{x}{2}}}{{1 - \cos \frac{x}{2}}}} } \right)

is equal to

View answer and explanation

Correct option: A

\frac{{ - 1}}{4}

Option A (

\frac{{ - 1}}{4}

) is correct because after simplifying the expression inside the arctangent function, the derivative is a constant.

243 Medium

243 of 932

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.

244 Medium

244 of 932

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.

245 Easy

245 of 932

Analyzing +4 | -1

y=x+e^x

, then

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

is equal to

View answer and explanation

Correct option: C

-\frac{e^x}{\left(1+e^x\right)^3}

Option C (-

\frac{e^x}{(1+e^x)^3}

) is correct because it is the result of finding the second derivative of the inverse function x(y).

246 Easy

246 of 932

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 evaluating the derivative of the function at

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

yields this result.

247 Medium

247 of 932

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.

248 Easy

248 of 932

Application +4 | -1

f(x)=\log |x|, x \neq 0

then

f^{\prime}(x)

equals

View answer and explanation

Correct option: B

\frac{1}{x}

Option B is correct because the derivative of log|x| is 1/x for all x not equal to 0.

249 Medium

249 of 932

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.

250 Medium

250 of 932

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.

251 Easy

251 of 932

Application +4 | -1

The differential coefficient of

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

with respect to

x

is:

View answer and explanation

Correct option: A

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

Option A (-1/(2 sqrt(1-x²))) is correct because it is the derivative of the given function after simplification using trigonometric identities.

252 Easy

252 of 932

Analyzing +4 | -1

{\cos ^{ - 1}}\left( {\frac{y}{2}} \right) = {\log _e}{\left( {\frac{x}{5}} \right)^5},\left| y \right| < 2,

then:

View answer and explanation

Correct option: D

{x^2}y'' + xy' + 25y = 0

Option D (

{x^2}y'' + xy' + 25y = 0

) is correct because after differentiating the given equation twice and simplifying, it satisfies this relation.

253 Easy

253 of 932

Analyzing +4 | -1

y = \frac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }},

then

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

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}\mathrm{cose}{\mathrm{c}^2}\frac{x}{2}

Option A (

\frac{1}{2}\mathrm{cose}{\mathrm{c}^2}\frac{x}{2}

) is correct because after simplifying the expression for y, its derivative is

\frac{1}{2}\mathrm{cose}{\mathrm{c}^2}\frac{x}{2}

254 Easy

254 of 932

Application +4 | -1

f\left( {x,y} \right) = \frac{{\cos (x - 4y)}}{{\cos \left( {x + 4y} \right)}},

then

{\left. {\frac{{\partial f}}{{\partial x}}} \right|_{y = \frac{\pi }{2}}}

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because when y = π/2, the function simplifies to 1, and its derivative with respect to x is 0.

255 Easy

255 of 932

Understanding +4 | -1

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

where

a,b,c

are parameters, then

y'

is equal to

View answer and explanation

Correct option: B

y'

Option B (

y'

) is correct because the third derivative of

y

is equal to the first derivative

y'

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256 Easy

256 of 932

Application +4 | -1

f\left( x \right) = x{\tan ^{ - 1}}x,

then

f'\left( 1 \right)

is equal to

View answer and explanation

Correct option: A

\frac{1}{2} + \frac{\pi }{4}

Option A (

\frac{1}{2} + \frac{\pi }{4}

) is correct because after finding the derivative of

f(x) = x\tan^{-1}x

and substituting

x = 1

, we get

f'(1) = \frac{1}{2} + \frac{\pi }{4}

257 Medium

257 of 932

Application +4 | -1

x=3 \cos t, y=4 \sin t

, then

d y / d x

at the point

(x=3 \sqrt{2} / 2, y=2 \sqrt{2})

View answer and explanation

Correct option: D

-\frac{4}{3}

Option D (

\frac{{-4}}{{3}}

) is correct because it is the value of

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

calculated at the given point after finding the derivatives.

258 Medium

258 of 932

Analyzing +4 | -1

Let

f

be twice differentiable function such that

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

and

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

h\left( x \right) = {\left\{ {f\left( x \right)} \right\}^2} + {\left\{ {\mathrm{g}\left( x \right)} \right\}^2}.

h\left( 5 \right) = 11,

then

h\left( {10} \right)

is equal to

View answer and explanation

Correct option: B

Option B (11) is correct because h(x) is a constant function, so h(5) = h(10) = 11.

259 Easy

259 of 932

Application +4 | -1

y = \sqrt {x{{\log }_e}x,}

, then

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

x = e

View answer and explanation

Correct option: B

\frac{1}{{\sqrt e }}

Option B (

\frac{1}{{\sqrt e }}

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

y = \sqrt {x{{\log }_e}x}

evaluated at

x = e

260 Medium

260 of 932

Analyzing +4 | -1

2y = {\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{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: B

x - \frac{\pi }{6}

Option B (

x - \frac{\pi}{6}

) is correct because after simplifying the given expression using trigonometric identities and inverse cotangent properties, the derivative is found to be

x - \frac{\pi}{6}

261 Hard

261 of 932

Analyzing +4 | -1

Let

f(x)=3 x^2+4 x g^{\prime}(1)+g^{\prime \prime}(2)

and,

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

for all

x \in R

. Then,

View answer and explanation

Correct option: D

All the above

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

262 Medium

262 of 932

Application +4 | -1

\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}}

263 Medium

263 of 932

Analyzing +4 | -1

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.

264 Easy

264 of 932

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

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

y

found by recognizing the infinite nested radical and using implicit differentiation.

265 Easy

265 of 932

Application +4 | -1

y = \frac{{1 + {x^2} + {x^4}}}{{1 + x + {x^2}}}\mathrm{ \,\,and \,\,}\frac{{dy}}{{dx}} = ax + b,

then

View answer and explanation

Correct option: C

a = 2,\,\,b = - 1

Option C (

a = 2,\,\,b = - 1

) is correct because after simplifying the given expression and differentiating, the coefficients of x and the constant term match these values.

266 Medium

266 of 932

Application +4 | -1

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.

267 Easy

267 of 932

Understanding +4 | -1

The derivative of

\log \left| x \right|

View answer and explanation

Correct option: C

\frac{1}{x},x \ne 0

Option C (

\frac{1}{x}, x \ne 0

) is correct because the derivative of

\log |x|

\frac{1}{x}

for both

x > 0

and

x < 0

268 Medium

268 of 932

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

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.

269 Easy

269 of 932

Application +4 | -1

If for

x \in\left(0, \frac{1}{4}\right)

, the derivative of

\tan ^{-1}\left(\frac{6 x \sqrt{x}}{1-9 x^3}\right)

\sqrt{x} \,g(x)

, then

g(x)

equals

View answer and explanation

Correct option: A

\frac{9}{1+9 x^3}

Option A (

\frac{9}{1+9 x^3}

) is correct because it is the result of differentiating the given function and simplifying the expression.

270 Easy

270 of 932

Application +4 | -1

f(x)=x^n+4

then the value of

f(1)+

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

View answer and explanation

Correct option: B

2^n+4

Option B (

2^n+4

) is correct because the expression is a Taylor series expansion of

f(x)

around

x=1

, which evaluates to

f(2) = 2^n + 4

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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