If $y^2+x^2=R^2$ and $k=1 / R$, then $k$ is equal to

$\frac{\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{2\left|y^{\prime \prime}\right|}{\sqrt{1+y^{\prime 2}}}$

$\frac{\left|y^{\prime \prime}\right|}{2 \sqrt{\left(1+y^{\prime 2}\right)^3}}$.

If $ \sin (x y)+\frac{x}{y}=x^{2}-y $, then $ \frac{d y}{d x} $ is equal to

$ \frac{\left[2 x y-y^{2} \cos (x y)-1\right]}{x y^{2} \cos (x y)+y^{2}-x} $

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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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691 Medium

691 of 932

Application +4 | -1

Let

f

and

g

be differentiable functions on

R

(the set of all real numbers) such that

g(1)=2=g^{\prime}(1)

and

f^{\prime}(0)=4

. If

h(x)=f(2 x g(x)+\cos \pi x-3)

then

h^{\prime}(1)

is equal to

View answer and explanation

Correct option: C

Option C (32) is correct because it is the value of

h'(1)

found by applying the chain rule and substituting the given values.

692 Medium

692 of 932

Application +4 | -1

f(2)=4

and

f^{\prime}(2)=1

, then

\lim\limits_{x \to 2} \frac{{x\,f(2) - 2f(x)}}{{x - 2}} =

View answer and explanation

Correct option: A

Option A (2) is correct because it is the result of evaluating the limit using the definition of the derivative and the given values.

693 Medium

693 of 932

Application +4 | -1

y = {a^x}.{b^{2x - 1}},

then

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

View answer and explanation

Correct option: D

y{(\log a{b^2})^2}

Option D (

y{(\log a{b^2})^2}

) is correct because it is the result of taking the second derivative of the given function using logarithmic differentiation.

694 Easy

694 of 932

Analyzing +4 | -1

f\left( {x,y} \right) = 2{\left( {x - y} \right)^2} - {x^4} - {y^4},

then

{\left| {\left( {{f_{xx}}{f_{yy}} - f_{xy}^2} \right)} \right|_{0,0}}

View answer and explanation

Correct option: C

Option C (0) is correct because after calculating the second-order partial derivatives and substituting (0,0), the expression evaluates to 0.

695 Medium

695 of 932

Application +4 | -1

Let

f(x)=1+\sin (3 x) \cdot g(x)

where

g(x)

is differentiable, then

f^{\prime}(0)

View answer and explanation

Correct option: B

3 g(0)

Option B (

3 g(0)

) is correct because the derivative

f'(x)

is calculated using the product rule and then evaluated at

x=0

696 Easy

696 of 932

Application +4 | -1

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it results from simplifying the expression using trigonometric identities and then differentiating.

697 Easy

697 of 932

Analyzing +4 | -1

y = \frac{{(a - x)\sqrt {a - x} - (b - x)\sqrt {x - b} }}{{\sqrt {a - x} + \sqrt {x - b} }}

, then find

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

whenever defined.

View answer and explanation

Correct option: C

\frac{{2x - (a + b)}}{{2\sqrt {(a - x)(x - b)} }}

Option C (

\frac{{2x - (a + b)}}{{2\sqrt {(a - x)(x - b)} }}

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

698 Easy

698 of 932

Application +4 | -1

Derivative of

{\log _{10}}x

with respect to

{x^2}

View answer and explanation

Correct option: B

\frac{{{{\log }_{10}}e}}{{2{x^2}}}

Option B (

\frac{{{{\log }_{{10}}}e}}{{2{{x^2}}}}

) is correct because it is the result of differentiating log base 10 of x with respect to x squared using the chain rule.

699 Easy

699 of 932

Application +4 | -1

The value of

\frac{d}{d x}(|x-1|+|x-5|)

x=3

View answer and explanation

Correct option: B

Option B (0) is correct because at

x=3

, the function

f(x) = |x-1| + |x-5|

is constant, hence its derivative is zero.

700 Medium

700 of 932

Analyzing +4 | -1

The value of

\log _e 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)

x=\frac{\pi}{4}

View answer and explanation

Correct option: D

4

Option D (4) is correct because it involves finding the derivative of the given logarithmic expression and evaluating it at

x=\frac{\pi}{4}

701 Medium

701 of 932

Application +4 | -1

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

and

f^{\prime}(x)=\sin x^{2}

, then

\frac{d y}{d x}

View answer and explanation

Correct option: C

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

The correct answer is C (

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

) because it is the derivative of y with respect to x using the chain rule.

702 Easy

702 of 932

Analyzing +4 | -1

Let

f\left( x \right) = {\left( {{x^3} + 2} \right)^{30}}

. If

{f^n}\left( x \right)

is a polynomial of degree 20, where

{f^n}\left( x \right)

denotes the

{n^{th}}

order derivative of

f\left( x \right)

with respect to

x

, then the value of

n

View answer and explanation

Correct option: C

Option C (70) is correct because the degree of the nth derivative is 90-n, and we want this to be 20, so n=70.

703 Hard

703 of 932

Application +4 | -1

The value of the derivative of

\left| {x - 1} \right| + \left| {x - 3} \right|

x = 2

View answer and explanation

Correct option: C

Option C (0) is correct because at x=2, the function is constant, and the derivative of a constant is zero.

704 Medium

704 of 932

Application +4 | -1

y^2+x^2=R^2

and

k=1 / R

, then

k

is equal to

View answer and explanation

Correct option: B

\frac{\left|y^{\prime \prime}\right|}{\sqrt{\left(1+y^{\prime 2}\right)^3}}

Option B (

\frac{{\left|y^{\prime \prime}\right|}}{{\sqrt{{\left(1+y^{\prime 2}\right)^3}}}}

) is correct because it represents the curvature (and hence the reciprocal of the radius) of the given circle.

705 Medium

705 of 932

Application +4 | -1

f(x)=\sin \left\{\frac{\pi}{3}[x]-x^2\right\}

for

2 < x < 3

and

[x]

denotes the greatest integer less than or equal to

x

, then

f'(\sqrt {\pi /3} )

is equal to

View answer and explanation

Correct option: B

-\sqrt{\pi / 3}

Option B (-

\sqrt{\pi / 3}

) is correct because it is the value of the derivative of the given function at the specified point, calculated using the chain rule.

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706 Medium

706 of 932

Analyzing +4 | -1

Let a function

y = y\left( x \right)

be defined parametrically by

x = 2t - \left| t \right|,y = {t^2} + t\left| t \right|

. Then,

y'\left( x \right),x > 0

View answer and explanation

Correct option: D

Does not exist

Option D (Does not exist) is correct because the derivative does not exist at x=0 due to different limits from the left and right.

707 Easy

707 of 932

Application +4 | -1

The left-hand derivative of

f(x)=[x] \sin \pi x

x=k

k

being an integer, is

View answer and explanation

Correct option: A

(-1)^k(k-1) \pi

Option A (

(-1)^k(k-1) \pi

) is correct because the left-hand derivative is computed using the limit definition and properties of the sine function near integer values.

708 Easy

708 of 932

Application +4 | -1

y = a\cos (\log x) + b\sin (\log x)

where

a,b

are parameters, then

{x^2}y'' + xy'

is equal to

View answer and explanation

Correct option: B

- y

Option B (

-y

) is correct because after differentiating twice and substituting, the expression

{x^2}y'' + xy'

simplifies to

-y

709 Easy

709 of 932

Application +4 | -1

y=\log x \cdot e^{\left(\tan x+x^{2}\right)}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: C

e^{\left(\tan x+x^{2}\right)}\left[\frac{1}{x}+\left(\sec ^{2} x+2 x\right) \log x\right]

Option C is correct because the derivative is found using the product rule and simplifying the resulting expression.

710 Medium

710 of 932

Application +4 | -1

f(x)=\left\{\begin{array}{ll}e^x, & x<2 \\ a x+b, & x \geq 2\end{array}\right.

is differentiable for all

x \in R

, then

View answer and explanation

Correct option: A

a=e^2, b=-e^2

Option A is correct because for the function to be differentiable at

x=2

, both continuity and differentiability must hold, leading to

a=e^2

and

b=-e^2

711 Easy

711 of 932

Application +4 | -1

u = {x^2} + {y^2}

and

x = s + 3t,y = 2s - t,

then

\frac{{{d^2}u}}{{d{s^2}}}

is equal to

View answer and explanation

Correct option: D

Option D (10) is correct because it is the result of applying the chain rule and differentiating twice with respect to s.

712 Easy

712 of 932

Application +4 | -1

y=\log _{e^x}(x-2)^2

for

x \neq 0,2

, then

y^{\prime}(3)

is equal to

View answer and explanation

Correct option: B

2 / 3

Option B (2/3) is correct because after applying the logarithmic and derivative rules, substituting x=3 gives 2/3.

713 Medium

713 of 932

Application +4 | -1

f(x)=\cos x \cos 2 x \cos 2^2 x \cos 2^3 x \ldots \ldots

\cos 2^{n-1} x

and

n>1

, then

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

View answer and explanation

Correct option: A

Option A (1) is correct because by using the identity

\sin(2x) = 2\sin(x)\cos(x)

repeatedly, we can simplify

f(x)

\frac{{\sin(2^n x)}}{{2^n \sin(x)}}

, then differentiate and evaluate at

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

714 Medium

714 of 932

Analyzing +4 | -1

The set of points where the function

f\left( x \right) = \left| {x - 1} \right|{e^x}

is differentiable, is

View answer and explanation

Correct option: B

R - \left\{ 1 \right\}

Option B (

R - \\{1\\}

) is correct because the absolute value function

|x - 1|

is not differentiable at

x = 1

, making the product

f(x) = |x - 1|e^x

also non-differentiable at that point.

715 Easy

715 of 932

Analyzing +4 | -1

f\left( x \right) = \cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x,

then the value of

f'\left( {\frac{\pi }{4}} \right)

View answer and explanation

Correct option: B

\sqrt 2

Option B (

\sqrt 2

) is correct because after simplifying the function using trigonometric identities and then differentiating, the value of f'(

\frac{\pi}{4}

) is

\sqrt 2

716 Easy

716 of 932

Application +4 | -1

Let

(x, y)

be any point on the unit circle with centre at the origin, then

View answer and explanation

Correct option: B

y y^{\prime \prime}+y^{\prime 2}+1=0

Option B (

y y^{\prime \prime}+y^{\prime 2}+1=0

) is correct because it is the result of differentiating the equation of the unit circle twice.

717 Medium

717 of 932

Application +4 | -1

f(x)=1-x^{2 / 3}+\frac{16}{x}

, then

f^{\prime}(-8)

View answer and explanation

Correct option: C

\frac{1}{12}

Option C (

\frac{{1}}{{12}}

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

f(x)

evaluated at

x = -8

718 Medium

718 of 932

Application +4 | -1

\frac{d}{d x}\left[\frac{e^{a x}}{\sin (b x+c)}\right]

is equal to

View answer and explanation

Correct option: C

\frac{e^{a x}[a \sin (b x+c)-b \cos (b x+c)]}{\sin ^{2}(b x+c)}

Option C is correct because the derivative is found using the quotient rule and simplifying the resulting expression.

719 Medium

719 of 932

Application +4 | -1

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

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

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

Option A (

\frac{y[2 x y-y^{2} \cos (x y)-1]}{x y^{2} \cos (x y)+y^{2}-x}

) is correct because it is the result of implicit differentiation.

720 Easy

720 of 932

Application +4 | -1

\frac{d}{{dx}}{\left( {\log x} \right)^4}

is equal to

View answer and explanation

Correct option: C

\frac{{4{{\left( {\log x} \right)}^3}}}{x}

Option C (

\frac{{4{{\left( {\log x} \right)}^3}}}{x}

) is correct because it is the result of applying the chain rule to differentiate (log x)⁴.

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