If $x{\log _e}\left( {{{\log }_e}x} \right) - {x^2} + {y^2} = 4\left( {y > 0} \right)\,$, then $\frac{{dy}}{{dx}}$ at $x = e$ is equal to

$\frac{e}{{\sqrt {4 + {e^2}} }}$

$\frac{{\left( {1 + 2e} \right)}}{{\sqrt {4 + {e^2}} }}$

$\frac{{\left( {1 + 2e} \right)}}{{2\sqrt {4 + {e^2}} }}$

If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{x\log \cos x}}{{\log \left( {1 + {x^2}} \right)}},x \ne 0}\\ {0,x = 0} \end{array}} \right.$, then

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

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

Let $f: R \rightarrow R$ be a function such that $f(x + y) = f(x) + f(y){\rm{ }}\,{\rm{for}}\,{\rm{ all }}\,x,y \in R$ If $f(x)$ is differentiable at $x=0$, then which one of the following is incorrect?

$f(x)$ is continuous for all $x \in R$

$f^{\prime}(x)$ is constant for all $x \in R$

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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451 Easy

451 of 932

Analyzing +4 | -1

Let

f(x)=\sqrt{x-1}+\sqrt{x+24-10 \sqrt{x-1}}, 1 < x < 26

be a real valued function, then

f^{\prime}(x)

for

1 < x < 26

, is

View answer and explanation

Correct option: A

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

452 Easy

452 of 932

Application +4 | -1

x{\log _e}\left( {{{\log }_e}x} \right) - {x^2} + {y^2} = 4\left( {y > 0} \right)\,

, then

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

x = e

is equal to

View answer and explanation

Correct option: B

\frac{{\left( {2e - 1} \right)}}{{2\sqrt {4 + {e^2}} }}

Option B (

\frac{{\left( {2e - 1} \right)}}{{2\sqrt {4 + {e^2}} }}

) is correct because it is the value of the derivative at x=e, obtained by implicit differentiation and substitution.

453 Easy

453 of 932

Application +4 | -1

Let

f

and

g

be differentiable functions such that

f(3)=5, g(3)=7, f^{\prime}(3)=13, g^{\prime}(3)=6, f^{\prime}(7)=2

and

g^{\prime}(7)=0

. If

h(x)=(f \circ g)(x)

, then

h^{\prime}(3)

is equal to

View answer and explanation

Correct option: C

Option C (12) is correct because, by the chain rule,

h'(x) = f'(g(x)) * g'(x)

, so

h'(3) = f'(g(3)) * g'(3) = f'(7) * 6 = 2 * 6 = 12

454 Medium

454 of 932

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.

455 Medium

455 of 932

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).

456 Easy

456 of 932

Application +4 | -1

Let

g(x)

be an antiderivative for

f(x)

. Then

\ln \left(1+(g(x))^2\right)

is an antiderivative for

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because differentiating

\ln \left(1+(g(x))^2\right)

gives the expression.

457 Easy

457 of 932

Application +4 | -1

\frac{d}{d x} x^{\log x}=

View answer and explanation

Correct option: A

x^{\log x}\left(\frac{2 \log x}{x}\right)

Option A (

x^{\log x}(\frac{{2 \log x}}{{x}})

) is correct because it's the result of applying logarithmic differentiation to the given function.

458 Easy

458 of 932

Understanding +4 | -1

The number of inflection points of a function given by a third degree polynomial is exactly

View answer and explanation

Correct option: B

Option B (1) is correct because a third-degree polynomial has exactly one inflection point.

459 Medium

459 of 932

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)

460 Easy

460 of 932

Application +4 | -1

f(1) = 1,\,\,f'(1) = 3,

then the derivative of

f(f(f(x))) + {(f(x))^2}\,\,\mathrm{at }\,\,x = 1

View answer and explanation

Correct option: D

Option D (33) is correct because it is the result of applying the chain rule and the given values of

f(1)

and

f'(1)

to the derivative of the composite function.

461 Hard

461 of 932

Application +4 | -1

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

, then the value of

f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\cdots

+\frac{1}{n !} f^{n}(0)

View answer and explanation

Correct option: B

2^{n}

The correct answer is B (

2^{n}

) because the given expression is the binomial expansion of

(1+1)^n

462 Easy

462 of 932

Analyzing +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{x\log \cos x}}{{\log \left( {1 + {x^2}} \right)}},x \ne 0}\\ {0,x = 0} \end{array}} \right.

, then

View answer and explanation

Correct option: B

f\left( x \right)

is not continuous and differentiable at

x = 0

Option D (None of these) is incorrect. The function is continuous and differentiable at x=0. The solution steps show the derivative exists and is -1/2.

463 Medium

463 of 932

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

464 Medium

464 of 932

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

465 Easy

465 of 932

Analyzing +4 | -1

Let

f: R \rightarrow R

be a function such that

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

f(x)

is differentiable at

x=0

, then which one of the following is incorrect?

View answer and explanation

Correct option: D

f(x)

is differentiable only in a finite interval containing zero

Option D (

f(x)

is differentiable only in a finite interval containing zero) is correct because the given condition implies that

f(x)

is differentiable for all

x \in R

, and

f'(x)

is constant for all

x \in R

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466 Medium

466 of 932

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.

467 Easy

467 of 932

Application +4 | -1

\frac{d}{{dx}}\left\{ {{{\sin }^2}\left( {{{\cot }^{ - 1}}\sqrt {\frac{{1 - x}}{{1 + x}}} } \right)} \right\}

equals

View answer and explanation

Correct option: B

\frac{1}{2}

Option B (

\frac{1}{2}

) is correct because after simplifying the expression using trigonometric identities, the derivative is simply

\frac{1}{2}

468 Easy

468 of 932

Analyzing +4 | -1

(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}

, where

a>b>0

, then

\frac{d x}{d y}

\left(\frac{\pi}{4}, \frac{\pi}{4}\right)

is:

View answer and explanation

Correct option: C

\frac{a+b}{a-b}

Option C is correct because by differentiating the given equation implicitly and evaluating at the specified point, we find that dx/dy = (a+b)/(a-b).

469 Medium

469 of 932

Application +4 | -1

\ln (x+y)=2 x y

, then

y^{\prime}(0)=

View answer and explanation

Correct option: A

Option A is correct because it is the result of implicit differentiation and substitution to find y'(0).

470 Easy

470 of 932

Understanding +4 | -1

Let

f

be a function defined for every

x

, such that

f^{\prime \prime}

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

then

f(x)

is equal to

View answer and explanation

Correct option: C

\sin x

Option C (

\sin x

) is the correct solution because it satisfies the given differential equation and initial conditions.

471 Easy

471 of 932

Analyzing +4 | -1

\sqrt {1 - {x^6}} + \sqrt {1 - {y^6}} = a\left( {{x^3} - {y^3}} \right)

and

\frac{{dy}}{{dx}} = f\left( {x,y} \right)\sqrt {\frac{{1 - {y^6}}}{{1 - {x^6}}}}

, then

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by substituting

{x^3} = \sin A

and

{y^3} = \sin B

, differentiating, and simplifying, we find that

f(x,y) = \frac{{{x^2}}}{{{y^2}}}

472 Easy

472 of 932

Analyzing +4 | -1

Let

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {1,\forall x < 0}\\ {1 + \sin x,\forall 0 \le x \le \pi /2} \end{array}} \right.

, then what is the value of

f'\left( x \right)

x = 0

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the left-hand derivative (LHD) and right-hand derivative (RHD) at x=0 are not equal.

473 Medium

473 of 932

Understanding +4 | -1

Let

f\left( x \right)

be a function such that

f'\left( a \right) \ne 0

. Then,at

x = a,\;f\left( x \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because a function can have a minimum or maximum even if its derivative is not zero at a point (e.g., at a cusp or endpoint).

474 Medium

474 of 932

Application +4 | -1

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

, then at

x=0, \frac{d y}{d x}

is equal to

View answer and explanation

Correct option: B

Option B is correct because substituting x=0 into the differentiated equation yields dy/dx = 1.

475 Easy

475 of 932

Application +4 | -1

Let

f_1(x)=e^x

and

f_{n+1}(x)=e^{f_n(x)}

for any

n \geq 1, n \in N

. Then for any fixed

n

, the value of

\frac{d}{d x} f_n(x)

equals :

View answer and explanation

Correct option: B

f_n(x) f_{n-1}(x) \ldots \ldots f_2(x) f_1(x)

Option B (

f_n(x) f_{n-1}(x) \ldots f_2(x) f_1(x)

) is correct because it is the result of repeatedly applying the chain rule.

476 Easy

476 of 932

Application +4 | -1

y = {\tan ^{ - 1}}\left( {\frac{{a\cos x - b\sin x}}{{b\cos x + a\sin x}}} \right),

then

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

is equal to

View answer and explanation

Correct option: B

- 1

Option B (-1) is correct because by simplifying the expression inside the inverse tangent function and differentiating, we get a constant derivative of -1.

477 Medium

477 of 932

Application +4 | -1

y=(\tan x)^{(\tan x)^{\tan x}}

, then at

x=\frac{\pi}{4}

, the value of

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

Option C is correct because at x=pi/4, the derivative simplifies to 2 after substituting the value.

478 Easy

478 of 932

Analyzing +4 | -1

Let

f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {\frac{{1 - x}}{x}} } \right)} \right),\,\,0 < x < 1

. Then,

View answer and explanation

Correct option: C

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

The correct answer is C (

{(1 - x)^2}f'(x) + 2{(f(x))^2} = 0

) because it is the relationship derived after simplifying the function and finding its derivative.

479 Easy

479 of 932

Application +4 | -1

y=f\left(x^3\right), z=g\left(x^2\right), f^{\prime}(x)=\cos x

and

g^{\prime}(x)=\sin x

, then

\frac{d y}{d z}

View answer and explanation

Correct option: A

\frac{3 x}{2} \cos x^3 \operatorname{cosec} x^2

Option A is correct because it follows from applying the chain rule to find dy/dx and dz/dx and then dividing them.

480 Hard

480 of 932

Analyzing +4 | -1

Let

f(x)

and

g(x)

be two functions having finite non-zero

3 r d

order derivatives

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

and

g^{\prime \prime \prime}(x)

for all

x \in R

. If

f(x) g(x)=1

for all

x \in R

, then

\frac{f^{\prime \prime \prime}}{f^{\prime}}-\frac{g^{\prime \prime \prime}}{g^{\prime}}

is equal to

View answer and explanation

Correct option: B

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

Option B (3(

\frac{f^{\prime\prime}}{f}-\frac{g^{\prime\prime}}{g}

)) is correct because after differentiating the equation f(x)g(x) = 1 three times and rearranging, the expression simplifies to 3(

\frac{f^{\prime\prime}}{f}-\frac{g^{\prime\prime}}{g}

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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