Differentiation — JEE Main PYQs

59 previous year questions on Differentiation (Mathematics), newest first. Open any question for its complete step-by-step solution.

Let $f (x) = x^{3} + x^{2} f^{'} (1) + 2 x f^{''} (2) + f^{'''} (3)$ , $x \in R$ . Then the value of $f^{'} (5)$ is :
JEE Main 2026View solution →
If $y (x) = sin x 271 cos x 281 sin x + cos x + 1 271$ , $x \in R$ , then $\frac{d ^{2} y}{d x ^{2}} + y$ is equal to
JEE Main 2025View solution →
Let $f (x) = x^{5} + 2 e^{x /4}$ for all $x \in R$ . Consider a function $g (x)$ such that $(g \circ f) (x) = x$ for all $x \in R$ . Then the value of $8 g^{'} (2)$ is
JEE Main 2024View solution →
Let $f (x) = x^{5} + 2 x^{3} + 3 x + 1, x \in R$ , and $g (x)$ be a function such that $f (g (x)) = x$ for all $x \in R$ . Then $\frac{g ( 7 )}{g ^{'} ( 7 )}$ is equal to:
JEE Main 2024View solution →
If $y (θ) = \frac{2 c o s θ + c o s 2 θ}{c o s 3 θ + 4 c o s 2 θ + 5 c o s θ + 2}$ , then at $θ = \frac{π}{2}$ , $y^{''} + y^{'} + y$ is equal to
JEE Main 2024View solution →
Suppose for a function $h, h (0) = 0, h (1) = 1$ and $h^{'} (0) = h^{'} (1) = 2$ . If $g (x) = h (e^{x}) e^{h (x)}$ , then $g^{'} (0)$ is equal to
JEE Main 2024View solution →
Let $f (x) = ax^{3} + bx^{2} + cx + 41$ be such that
$f (1) = 40, f^{'} (1) = 2$ and $f^{''} (1) = 4$ .
Then $a^{2} + b^{2} + c^{2}$ is equal to :
JEE Main 2024View solution →
If $lo g_{e} y = 3 sin^{- 1} x$ , then $(1 - x)^{2} y^{''} - x y^{'}$ at $x = \frac{1}{2}$ is equal to :
JEE Main 2024View solution →
Let $y = lo g_{e} (\frac{1 - x ^{2}}{1 + x ^{2}}), - 1 < x < 1$ . Then at $x = \frac{1}{2}$ , the value of $225 (y^{'} - y^{''})$ is equal to
JEE Main 2024View solution →
If $f (x) = 2 cos^{4} x 3 + 2 cos^{4} x 2 cos^{4} x 2 sin^{4} x 2 sin^{4} x 3 + 2 sin^{4} x 3 + sin^{2} 2 x sin^{2} 2 x sin^{2} 2 x$ then $\frac{1}{5} f^{'} (0)$ is equal to
JEE Main 2024View solution →
Let $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3)$ , $x \in R$ . Then $f^{'} (10)$ is equal to ____.
JEE Main 2024View solution →
If $y = \frac{( x + 1 ) ( x ^{2} - x )}{x x + x + x} + \frac{1}{15} (3 cos^{2} x - 5) cos^{3} x$ , then $96 y^{'} (\frac{π}{6})$ is equal to :
JEE Main 2024View solution →
If $f (x) = x^{3} 3 x^{2} + 2 x^{3} - x 2 x^{2} + 1 2 x 4 1 + 3 x x^{3} + 6 x^{2} - 2$ for all $x \in R$ , then $2 f (0) + f^{'} (0)$ is equal to
JEE Main 2024View solution →
Suppose
$f (x) = \frac{( 2 ^{x} + 2 ^{- x} ) t a n x t a n ^{- 1} ( x ^{2} - x + 1 )}{( 7 x ^{2} + 3 x + 1 ) ^{3}}$ ,
Then the value of $f^{'} (0)$ is equal to
JEE Main 2024View solution →
Let $f (x) = \frac{s i n x + c o s x - 2}{s i n x - c o s x}, x \in [0, π] - {\frac{π}{4}}$ . Then $f (\frac{7 π}{12}) f^{''} (\frac{7 π}{12})$ is equal to
JEE Main 2023View solution →
If $2 x^{y} + 3 y^{x} = 20$ , then $\frac{d y}{d x}$ at $(2, 2)$ is equal to
JEE Main 2023View solution →
Let $f (x) = \sum_{k = 1}^{10} kx^{k}, x \in R$ . If $2 f (2) + f^{'} (2) = 119 (2)^{n} + 1$ then n is equal to ______.
JEE Main 2023View solution →
If $y (x) = x^{x}, x > 0$ , then $y^{''} (2) - 2 y^{'} (2)$ is equal to :
JEE Main 2023View solution →
If $f (x) = x^{3} - x^{2} f^{'} (1) + x f^{''} (2) - f^{'''} (3)$ , $x \in R$ , then
JEE Main 2023View solution →
Let
$y = f (x) = sin^{3} (\frac{π}{3} (cos (\frac{π}{3 2} (- 4 x^{3} + 5 x^{2} + 1)^{\frac{3}{2}})))$ .
Then, at $x = 1$ ,
JEE Main 2023View solution →
If $f (x) = x^{2} + g^{'} (1) x + g^{''} (2)$ and
$g (x) = f (1) x^{2} + x f^{'} (x) + f^{''} (x)$ ,
then the value of $f (4) - g (4)$ is equal to ____.
JEE Main 2023View solution →
Let
$y (x) = (1 + x) (1 + x^{2}) (1 + x^{4}) (1 + x^{8}) (1 + x^{16})$ .
Then $y^{'} - y^{''}$ at $x = - 1$ is equal to
JEE Main 2023View solution →
Let $x (t) = 22 cos t sin 2 t$ and
$y (t) = 22 sin t sin 2 t, t \in (0, \frac{π}{2})$ . Then $\frac{1 + ( \frac{d y}{d x} ) ^{2}}{\frac{d ^{2} y}{d x ^{2}}}$ at $t = \frac{π}{4}$ is equal to
JEE Main 2022View solution →
The value of $lo g_{e} 2 \frac{d}{d x} (lo g_{c o s x} cosec x)$ at $x = \frac{π}{4}$ is
JEE Main 2022View solution →
If $y (x) = (x^{x})^{x}, x > 0$ then $\frac{d ^{2} x}{dy ^{2}} + 20$ at $x = 1$ is equal to:
JEE Main 2022View solution →
If $y = tan^{- 1} (sec x^{3} - tan x^{3}), \frac{π}{2} < x^{3} < \frac{3 π}{2}$ , then
JEE Main 2022View solution →
If the angle made by the tangent at the point $(x_{0}, y_{0})$ on the curve $x = 12 (t + sin t cos t)$ , $y = 12 (1 + sin t)^{2}$ , $0 < t < \frac{π}{2}$ , with the positive $x$ -axis is $\frac{π}{3}$ , then $y_{0}$ is equal to
JEE Main 2022View solution →
Let $f : R \to R$ be defined as $f (x) = x^{3} + x - 5$ . If $g (x)$ is a function such that $f (g (x)) = x$ , $\forall x \in R$ , then $g^{'}$ (63) is equal to
JEE Main 2022View solution →
If the curve $y = a x^{2} + b x + c, x \in R$ passes through the point $(1, 2)$ and the tangent line to this curve at origin is $y = x$ , then the possible values of $a, b, c$ are:
JEE Main 2021View solution →
If $y (x) = cot^{- 1} (\frac{1 + s i n x + 1 - s i n x}{1 + s i n x - 1 - s i n x}), x \in (\frac{π}{2}, π)$ , then $\frac{d y}{d x}$ at $x = \frac{5 π}{6}$ is:
JEE Main 2021View solution →
Let $f (x)$ be a differentiable function at $x = a$ with $f^{'} (a) = 2$ and $f (a) = 4$ . Then $lim_{x \to a} \frac{x f ( a ) - a f ( x )}{x - a}$ equals :
JEE Main 2021View solution →
Let $f : R \to R$ be a function such that $f (2) = 4$ and $f^{'} (2) = 1$ . Then the value of $lim_{x \to 2} \frac{x ^{2} f ( 2 ) - 4 f ( x )}{x - 2}$ is equal to :
JEE Main 2021View solution →
If $f (x) = sin (cos^{- 1} (\frac{1 - 2 ^{2 x}}{1 + 2 ^{2 x}}))$ and its first derivative with respect to $x$ is $- \frac{b}{a} lo g_{e} 2$ when $x = 1$ , where $a$ and $b$ are integers, then the minimum value of $a^{2} - b^{2}$ is. ____
JEE Main 2021View solution →
Let $f (x) = x^{6} + 2 x^{4} + x^{3} + 2 x + 3, x \in R$ . Then the natural number $n$ for which $lim_{x \to 1} \frac{x ^{n} f ( 1 ) - f ( x )}{x - 1} = 44$ is ______.
JEE Main 2021View solution →
If $y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}, α \in (\frac{3 π}{4}, π)$ then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is:
JEE Main 2020View solution →
Let $x^{k} + y^{k} = a^{k}, (a, k > 0)$ and $\frac{d y}{d x} + (\frac{y}{x})^{\frac{1}{3}} = 0$ , then $k$ is:
JEE Main 2020View solution →
Let $f$ and $g$ be differentiable functions on $R$ such that fog is the identity function. If for some $a, b \in R, g^{'} (a) = 5$ and $g (a) = b$ , then $f^{'} (b)$ is equal to:
JEE Main 2020View solution →
Which of the following point lies on the tangent to the curve $x^{4} e^{y} + 2 y + 1 = 3$ at the point $(1, 0)$ ?
JEE Main 2020View solution →
The derivation of $tan^{- 1} (\frac{1 + x ^{2} - 1}{x})$ with respect to $tan^{- 1} (\frac{2 x 1 - x ^{2}}{1 - 2 x ^{2}})$ at $x = \frac{1}{2}$ is:
JEE Main 2020View solution →
If $y^{2} + lo g_{e} (cos^{2} x) = y, x \in (- \frac{π}{2}, \frac{π}{2})$ ,

Then:
JEE Main 2020View solution →
If $2 y = (cot^{- 1} (\frac{3 c o s x + s i n x}{c o s x - 3 s i n x}))^{2}, x \in (0, \frac{π}{2})$ then $\frac{d y}{d x}$ is equal to
JEE Main 2019View solution →
If $f (1) = 1, f^{'} (1) = 3$ , then the derivative of $f (f (f (x))) + (f (x))^{2}$ at $x = 1$ is:
JEE Main 2019View solution →
If the tangent to the curve, $y = x^{3} + a x - b$ at the point ( $1, - 5$ ) is perpendicular to the line, $- x + y + 4 = 0$ , then which one of the following, points lies on the curve?
JEE Main 2019View solution →
Let $f : R \to R$ be a function such that $f (x) = x^{3} + x^{2} f^{'} (1) + x f^{''} (2) + f^{'''} (3), x \in R$ . Then $f (2)$ equals:
JEE Main 2019View solution →
If $e^{y} + x y = e$ , the ordered pair $(\frac{d y}{d x}, \frac{d ^{2} y}{d x ^{2}})$ at $x = 0$ is equal to :
JEE Main 2019View solution →
Let $f (x) = lo g_{e} (sin x), (0 < x < π)$ and $g (x) = sin^{- 1} (e^{- x}), (x \geq 0)$ . If $α$ is a positive real number such that $a = (fog)^{'} (α)$ and $b = (fog) (α)$ , then
JEE Main 2019View solution →
If $x lo g_{e} (lo g_{e} x) - x^{2} + y^{2} = 4 (y > 0)$ , then $\frac{d y}{d x}$ at $x = e$ is equal to:
JEE Main 2019View solution →
For $x > 1$ , if $(2 x)^{2 y} = 4 e^{2 x - 2 y}$ , then $(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}$ is equal to:
JEE Main 2019View solution →
The derivative of $tan^{- 1} (\frac{s i n x - c o s x}{s i n x + c o s x})$ , with respect to $\frac{x}{2}$ , where $(x \in (0, \frac{π}{2}))$ is:
JEE Main 2019View solution →
Let $f$ be a differentiable function such that $f (1) = 2$ and $f^{'} (x) = f (x)$ for all $x \in R$ . If $h (x) = f (f (x))$ , then $h^{'} (1)$ is equal to :
JEE Main 2019View solution →