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JEE • MATHEMATICS • DIFFERENTIATION

Differentiation practice · Mixed difficulty

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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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811 Easy
811 of 932
Application +4 | -1
If y=log⁡cos⁡xsin⁡xy=\log _{\cos x} \sin x, then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: A
(cot⁡xlog⁡cos⁡x+tan⁡xlog⁡sin⁡x)/(log⁡cos⁡x)2(\cot x \log \cos x+\tan x \log \sin x) /(\log \cos x)^2
Option A is correct because it is the result of applying the quotient rule and chain rule to differentiate the given logarithmic function.
812 Medium
812 of 932
Understanding +4 | -1
Assertion : ddxecos⁡x=ecos⁡x(−sin⁡x)\frac{d}{d x} e^{\cos x}=e^{\cos x}(-\sin x)

Reason: ddxex=ex\frac{d}{d x} e^x=e^x
View answer and explanation
Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because both the assertion and the reason are correct statements, but the reason does not provide a direct explanation for the assertion.
813 Medium
813 of 932
Application +4 | -1
If f(x)=∣log⁡e∣x∣∣f\left( x \right) = \left| {{{\log }_e}\left| x \right|} \right|, then f′(x)f'\left( x \right) equals
View answer and explanation
Correct option: B
1x\frac{1}{x} for ∣x∣>1\left| x \right| > 1 and −1x\frac{{ - 1}}{x} for ∣x∣<1\left| x \right| < 1
Option B (1x\frac{1}{x} for ∣x∣>1\left| x \right| > 1 and −1x\frac{{ - 1}}{x} for ∣x∣<1\left| x \right| < 1) is correct because it correctly describes the derivative of the given function for different ranges of x.
814 Medium
814 of 932
Application +4 | -1
If gg is the inverse of a function ff and f′(x)=11+x5f^{\prime}(x)=\frac{1}{1+x^5}, then g′(x)g^{\prime}(x) is equal to
View answer and explanation
Correct option: D
1+{g(x)}51+\{g(x)\}^5
Option D (1+{g(x)}51+\{g(x)\}^5) is correct because it is derived from differentiating the inverse function relationship and substituting the given derivative of f(x).
815 Medium
815 of 932
Application +4 | -1
If f(x)=log⁡x′(log⁡x)f(x)=\log _{x^{\prime}}(\log x), then f′(x)f^{\prime}(x) at x=ex=e is
View answer and explanation
Correct option: D
12e\frac{1}{2 e}
Option D (12e\frac{1}{2 e}) is correct because after simplifying the function and finding its derivative, the value of f′(e)f'(e) is 12e\frac{1}{2e}.
816 Easy
816 of 932
Application +4 | -1
If f(x)=(1+x)nf(x)=(1+x)^n, then the value of f(0)+f′(0)+f′′(0)2!+…+fn(0)n!f(0)+f^{\prime}(0)+\frac{f^{\prime \prime}(0)}{2 !}+\ldots+\frac{f^n(0)}{n !} is
View answer and explanation
Correct option: B
2n2^n
Option B (2n2^n) is correct because the given expression is the binomial expansion of (1+1)n(1+1)^n evaluated at x=0x=0.
817 Medium
817 of 932
Application +4 | -1
If f(x)=1+nx+n(n−1)2x2+n(n−1)(n−2)6x3+.…+xn,f\left( x \right) = 1 + nx + \frac{{n\left( {n - 1} \right)}}{2}{x^2} + \frac{{n\left( {n - 1} \right)\left( {n - 2} \right)}}{6}{x^3} + . \ldots + {x^n}, then, f′′(1)f''\left( 1 \right) is equal to
View answer and explanation
Correct option: C
n(n−1)2n−2n\left( {n - 1} \right){2^{n - 2}}
Option C (n(n−1)2n−2n(n - 1)2^{n - 2}) is correct because the given series is the binomial expansion of (1+x)n(1 + x)^n, and its second derivative evaluated at x=1 is n(n−1)2n−2n(n - 1)2^{n - 2}.
818 Medium
818 of 932
Analyzing +4 | -1
If x2+y2=a2{x^2} + {y^2} = {a^2} and k=1/ak = 1/a, then kk is equal to
View answer and explanation
Correct option: B
∣y′′∣(1+y′2)3\frac{{\left| {y''} \right|}}{{\sqrt {{{\left( {1 + {{y'}^2}} \right)}^3}} }}
The correct answer is B (∣y′′∣(1+y′2)3\frac{{\left| {y''} \right|}}{{\sqrt {{{\left( {1 + {{y'}^2}} \right)}^3}} }}) because it represents the radius of curvature k=1ak = \frac{1}{a} derived from the given equation and its derivatives.
819 Medium
819 of 932
Analyzing +4 | -1
Let f(x)=x2+xg′(1)+g′′(2)f(x)=x^2+x g^{\prime}(1)+g^{\prime \prime}(2) and g(x)=x2+xf′(2)+f′′(3)g(x)=x^2+x f^{\prime}(2)+f^{\prime \prime}(3). Then,
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all the given relationships between the derivatives of f(x) and g(x) are valid based on the given equations.
820 Medium
820 of 932
Application +4 | -1
Let yy be an implicit function of xx defined by x2x−2xxcot⁡y−1=0x^{2 x}-2 x^{x} \cot y-1=0. Then y′(1)y^{\prime}(1) equals
View answer and explanation
Correct option: A
-1
The correct answer is A (-1) because it is the value of the derivative y'(1) obtained after implicit differentiation and substitution.
821 Medium
821 of 932
Analyzing +4 | -1
If y=ax2(x−a)(x−b)(x−c)+bx(x−b)(x−c)+cx−c+1y=\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{c}{x-c}+1 then dy/dxd y / d x is equal to
View answer and explanation
Correct option: A
−yx(ax−a+bx−b+cx−c)\frac{-y}{x}\left(\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\right)
Option A is correct because the derivative is found by logarithmic differentiation, simplifying the expression before differentiating.
822 Medium
822 of 932
Analyzing +4 | -1
If y=xxxx…∞y = {x^{{x^x}{{^x}^{{ \ldots ^\infty }}}}}, then x(1−ylog⁡x)dydxx\left( {1 - y\log x} \right)\frac{{dy}}{{dx}}
View answer and explanation
Correct option: B
y2{y^2}
Option B (y2{y^2}) is correct because after taking logarithms and differentiating, the expression simplifies to y2y^2.
823 Easy
823 of 932
Understanding +4 | -1
For a differentiable function ff on R\mathbf{R} there is an x0x_0 with
f′(x0)=12[f′(0)+f′(1)]f^{\prime}\left(x_0\right)=\frac{1}{2}\left[f^{\prime}(0)+f^{\prime}(1)\right]
View answer and explanation
Correct option: D
if ff is continuously differentiable
Option D (if ff is continuously differentiable) is correct because the Intermediate Value Theorem can be applied to the derivative of a continuously differentiable function.
824 Easy
824 of 932
Application +4 | -1
If f(x)=x+tan⁡xf(x)=x+\tan x and g(x)g(x) is the inverse of f(x)f(x) then g′(x)g^{\prime}(x) is equal to
View answer and explanation
Correct option: C
12+[g(x)−x]2\frac{1}{2+[g(x)-x]^2}
Option B (12+[g(x)−x]2\frac{{1}}{{2+[g(x)-x]^2}}) is correct because by differentiating f(g(x))=xf(g(x)) = x and using trigonometric identities, we arrive at the given expression.
825 Easy
825 of 932
Application +4 | -1
The point of inflection of y=x3−5x2+y=x^3-5 x^2+ 3x−53 x-5 is
View answer and explanation
Correct option: D
54\frac{5}{4}
Option D 54\frac{{5}}{{4}} is correct because it is the x-coordinate where the second derivative of the function is zero and the third derivative is non-zero.
826 Easy
826 of 932
Application +4 | -1
Let f(x+y)=f(x)f(y)f(x+y)=f(x) f(y) for all xx and yy, and f(5)=−2,f′(0)=3f(5)=-2, f^{\prime}(0)=3, then f′(5)f^{\prime}(5) is equal to :
View answer and explanation
Correct option: C
-6
Option C (-6) is correct because it is the value of f′(5)f'(5) found using the given functional equation and derivative at 0.
827 Easy
827 of 932
Application +4 | -1
If the equation 2sin⁡2x+sin⁡2x2=k2 \sin ^2 \mathrm{x}+\frac{\sin 2 \mathrm{x}}{2}=\mathrm{k} has at-least one real solution, then the sum of all integral values of k\mathrm{k} is
View answer and explanation
Correct option: A
3
Option A (3) is correct because the equation can be rewritten as a trigonometric function with a range that allows for only three integer values of k.
828 Easy
828 of 932
Understanding +4 | -1
If f:R→Rf:R \to R is an even function which us twice differentiable on RR and f′′(π)=1f''\left( \pi \right) = 1, then f′′(−π)f''\left( { - \pi } \right)
View answer and explanation
Correct option: C
1
Option C (1) is correct because the second derivative of an even function is also even, hence f′′(−x)=f′′(x)f''(-x) = f''(x).
829 Medium
829 of 932
Analyzing +4 | -1
If f(x)={xe−(1∣x∣+1x),x≠00,x=0f(x)=\left\{\begin{array}{ll}x e^{-\left(\frac{1}{|x|}+\frac{1}{x}\right)} & , x \neq 0 \\ 0 & , x=0\end{array}\right. then f(x)\mathrm{f}(\mathrm{x}) is
View answer and explanation
Correct option: C
continuous for all xx but not differentiable at x=0x=0
Option C is correct because the function is continuous for all x but not differentiable at x=0 due to unequal left-hand and right-hand derivatives.
830 Easy
830 of 932
Application +4 | -1
Let f(x)=x2−xx2+2x,x≠0,−2f(x)=\frac{x^2-x}{x^2+2 x}, x \neq 0,-2. Then ddx(f−1(x))\frac{d}{d x}\left(f^{-1}(x)\right) (wherever defined) is equal to
View answer and explanation
Correct option: B
3(1−x)2\frac{3}{(1-x)^2}
Option B (3(1−x)2\frac{3}{(1-x)^2}) is correct because it is the derivative of the inverse function after simplification.
831 Easy
831 of 932
Analyzing +4 | -1
Let f(x)=[∣x∣]f(x)=[|x|] where [.] denotes the greatest integer function, then f′(−1)f^{\prime}(-1) is
View answer and explanation
Correct option: C
non-existent
Option C (non-existent) is correct because the left-hand derivative and right-hand derivative at x=-1 are not equal, making the function non-differentiable.
832 Medium
832 of 932
Application +4 | -1
If f′(x)=sin⁡2xf^{\prime}(x)=\sin ^2 x and
y=f(2x−1x2+1)y=f\left(\frac{2 x-1}{x^2+1}\right)
then dy/dxd y / d x at x=1x=1 is
View answer and explanation
Correct option: D
12sin⁡2(12)\frac{1}{2} \sin ^2\left(\frac{1}{2}\right)
Option D is correct because it is the result of applying the chain rule and evaluating the derivative at x=1.
833 Hard
833 of 932
Analyzing +4 | -1
If f(x)=x3+x2f′(1)+xf′′(2)+f′′′(3)f(x)=x^3+x^2 f^{\prime}(1)+x f^{\prime \prime}(2)+f^{\prime \prime \prime}(3) for all x∈Rx \in R, then which of the following is false?
View answer and explanation
Correct option: D
f(1)+f(3)=f(0)f(1)+f(3)=f(0)
Option D, f(1)+f(3)=f(0)f(1)+f(3)=f(0), is false because the correct relationship between the function values is f(0)+f(2)=f(1)f(0)+f(2)=f(1).
834 Medium
834 of 932
Application +4 | -1
If y=1+log⁡xx−xlog⁡xy=\frac{1+\log x}{x-x \log x}, then dy/dxd y / d x is equal to
View answer and explanation
Correct option: B
x2y2+12x2\frac{x^2 y^2+1}{2 x^2}
The correct answer is A (x2y2+12x2\frac{{x^2 y^2+1}}{{2 x^2}}) because after differentiating the given equation implicitly and substituting the value of log⁡x\log x from the original equation, we arrive at this expression.
835 Medium
835 of 932
Application +4 | -1
Let ef(x)=log⁡x.{e^{f(x)}} = \log x. If g(x)g(x) is the inverse of f(x),f(x), then find g′(x).g'(x).
View answer and explanation
Correct option: C
eex+x{e^{{e^x} + x}}
Option A (eex{e^{{e^x}}}) is incorrect. Option B (ex{e^x}) is incorrect. Option C (eex+x{e^{{e^x} + x}}) is correct because it is the derivative of the inverse function g(x) found by using the relationship between f(x) and g(x). Option D (None of these) is incorrect.
836 Medium
836 of 932
Application +4 | -1
If xy.yx=16{x^y}.{y^x} = 16, then the value of dydx\frac{{dy}}{{dx}} at (2,2) is
View answer and explanation
Correct option: A
−1- 1
Option A (- 1) is correct because after taking the logarithm and differentiating implicitly, the derivative at the point (2, 2) is -1.
837 Medium
837 of 932
Understanding +4 | -1
If f(x)=cos⁡−1{1+(log⁡ex)21+(log⁡ex)2}f\left( x \right) = {\cos ^{ - 1}}\left\{ {\frac{{1 + {{\left( {{{\log }_e}x} \right)}^2}}}{{1 + {{\left( {{{\log }_e}x} \right)}^2}}}} \right\}, then f′(e)f'\left( e \right)
View answer and explanation
Correct option: C
Is equal to 1e\frac{1}{e}
Option C (Is equal to 1e\frac{1}{e}) is correct because after simplification and differentiation, substituting x = e yields the value 1e\frac{1}{e}.
838 Easy
838 of 932
Application +4 | -1
Let F(x)=f(x)g(x)h(x)\mathrm{F(x)=f(x) g(x) h(x)} for all real x\mathrm{x}, where f(x),g(x)\mathrm{f(x), g(x)} and h(x)\mathrm{h}(\mathrm{x}) are differentiable functions. At some point x0,F′(x0)=21F(x0),f′(x0)=4f(x0),g′(x0)=−7g(x0)\mathrm{x}_0, \mathrm{F}^{\prime}\left(\mathrm{x}_0\right)=21 \mathrm{F}\left(\mathrm{x}_0\right), \mathrm{f}^{\prime}\left(\mathrm{x}_0\right)=4 \mathrm{f}\left(\mathrm{x}_0\right), \mathrm{g}^{\prime}\left(\mathrm{x}_0\right)=-7 \mathrm{g}\left(\mathrm{x}_0\right) and h′(x0)=kh⁡(x0)\mathrm{h}^{\prime}\left(\mathrm{x}_0\right)=\operatorname{kh}\left(\mathrm{x}_0\right). Then the value of k/8\mathrm{k} / 8 is
View answer and explanation
Correct option: A
3
Option A (3) is correct because by differentiating F(x) and substituting the given conditions, the value of k/8 is found to be 3.
839 Easy
839 of 932
Application +4 | -1
If y=cos⁡−1(4cos⁡x−5sin⁡x41)y=\cos ^{-1}\left(\frac{4 \cos x-5 \sin x}{\sqrt{41}}\right), then dydx\frac{d y}{d x} is equal to
View answer and explanation
Correct option: B
1
Option B (1) is correct because the function simplifies to y = x + constant, making the derivative equal to 1.
840 Medium
840 of 932
Application +4 | -1
Suppose ff is a differentiable real function such that f(x)+f′(x)≤1f(x)+f^{\prime}(x) \leq 1 for all xx and f(0)=0f(0)=0, then the largest possible value of f(1)f(1) is
View answer and explanation
Correct option: C
1−e−11-e^{-1}
The correct answer is C (1−e−11-e^{-1}) because it is the largest possible value of f(1) derived from the given differential inequality and initial condition.
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