Techniques of Differentiation

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For $x \in R, f (x) = ∣ lo g 2 - sin x ∣$ and $g (x) = f (f (x))$ , then

g'(0) = -\cos(\log 2)

g is differentiable at x = 0 and g'(0) = -\sin(\log 2)

g is not differentiable at x = 0

g'(0) = \cos(\log 2)

If $g$ is the inverse of a function $f$ and $f^{'} (x) = \frac{1}{1 + x ^{5}}$ , then $g^{'} (x)$ is equal to

\frac{1}{1 + { g ( x ) } ^{5}}

1 + {g (x)}^{5}

1 + x^{5}

5 x^{4}

If $y = sec (tan^{- 1} x)$ , then $\frac{d y}{d x}$ at $x = 1$ is equal to

\frac{1}{2}

\frac{1}{2}

1

2

Let $f : (- 1, 1) \to R$ be a differentiable function with $f (0) = - 1$ and $f^{'} (0) = 1$ . Let $g (x) = [f (2 f (x) + 2)]^{2}$ . Then $g^{'} (0) =$

- 4

0

- 2

4

Let $y$ be an implicit function of $x$ defined by $x^{2 x} - 2 x^{x} cot y - 1 = 0$ . Then $y^{'} (1)$ equals

1

lo g 2

- lo g 2

- 1

If $x^{m} \cdot y^{n} = (x + y)^{m + n}$ , then $\frac{d y}{d x}$ is

\frac{y}{x}

\frac{x + y}{x y}

x y

\frac{x}{y}

If $x = e^{y + e^{y + e^{y + \dots \infty}}}, x > 0$ , then $\frac{d y}{d x}$ is

\frac{1 + x}{x}

\frac{1}{x}

\frac{1 - x}{x}

\frac{x}{1 + x}

If $f (x + y) = f (x) f (y) \forall x, y$ and $f (5) = 2, f^{'} (0) = 3$ , then $f^{'} (5)$ is

JEE Advanced 2016

Let $f : R \to R, g : R \to R$ and $h : R \to R$ be differentiable functions such that $f (x) = x^{3} + 3 x + 2, g (f (x)) = x$ and $h (g (g (x))) = x$ for all $x \in R$ . Then

* Multiple Correct Options

g^{'} (2) = \frac{1}{15}

h^{'} (1) = 666

h (0) = 16

h (g (3)) = 36

JEE Advanced 2004

If $y$ is a function of $x$ and $lo g (x + y) - 2 x y = 0$ , then the value of $y^{'} (0)$ is equal to

1

- 1

2

0

JEE Advanced 1994

If $y = (sin x)^{t a n x}$ , then $\frac{d y}{d x}$ is equal to

(sin x)^{t a n x} (1 + sec^{2} x lo g sin x)

tan x (sin x)^{t a n x - 1} \cdot cos x

(sin x)^{t a n x} sec^{2} x lo g sin x

tan x (sin x)^{t a n x - 1}

JEE Advanced 1983

If $G (x) = - 25 - x^{2}$ then $lim_{x \to 1} \frac{G ( x ) - G ( 1 )}{x - 1}$ has the value

1/24

1/5

- 24

JEE Advanced 2014

The slope of the tangent to the curve $(y - x^{5})^{2} = x (1 + x^{2})^{2}$ at the point $(1, 3)$ is

JEE Advanced 2011

Let $f (θ) = sin (tan^{- 1} (\frac{s i n θ}{c o s 2 θ}))$ , where $- \frac{π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d ( t a n θ )} (f (θ))$ is

JEE Advanced 2009

If the function $f (x) = x^{3} + e^{x /2}$ and $g (x) = f^{- 1} (x)$ , then the value of $g^{'} (1)$ is

JEE Advanced 2003

If a function $f : [- 2 a, 2 a] \to R$ is an odd function such that $f (x) = f (2 a - x)$ for $x \in [a, 2 a]$ and the left hand derivative at $x = a$ is 0 then find the left hand derivative at $x = - a$ .

JEE Advanced 1996

If $x e^{x y} = y + sin^{2} x$ , then at $x = 0, \frac{d y}{d x} = \dots\dots\dots$

JEE Advanced 1991

LEVELJEE Advanced

Find $\frac{d y}{d x}$ at $x = - 1$ , when $(sin y)^{s i n (\frac{π x}{2})} + \frac{3}{2} sec^{- 1} (2 x) + 2^{x} tan (ln (x + 2)) = 0$ .

JEE Advanced 1990

If $f (x) = ∣ x - 2∣$ and $g (x) = f [f (x)]$ , then $g^{'} (x) = \dots\dots\dots$ for $x > 20$ .

JEE Advanced 1986

The derivative of $sec^{- 1} (\frac{1}{2 x ^{2} - 1})$ with respect to $1 - x^{2}$ at $x = \frac{1}{2}$ is $\dots\dots\dots$

JEE Advanced 2004

$f^{'} (0) = lim_{n \to \infty} n f (1/ n)$ and $f (0) = 0$ . Using this find $lim_{n \to \infty} ((n + 1) \frac{2}{π} cos^{- 1} (\frac{1}{n}) - n) \cdot cos^{- 1} \frac{1}{n} < \frac{π}{2} .$

JEE Advanced 1995

Let $f (\frac{x + y}{2}) = \frac{f ( x ) + f ( y )}{2}$ for all real $x$ and $y$ . If $f^{'} (0)$ exists and equals $- 1$ and $f (0) = 1$ , find $f (2)$ .

JEE Advanced 1990

A function $f : R \to R$ satisfies the equation $f (x + y) = f (x) f (y)$ for all $x, y$ in $R$ and $f (x) \neq = 0$ for any $x$ in $R$ . Let the function be differentiable at $x = 0$ and $f^{'} (0) = 2$ . Show that $f^{'} (x) = 2 f (x)$ for all $x$ in $R$ . Hence, determine $f (x)$ .

JEE Advanced 1985

If $f (x) = lo g_{x} (ln x)$ , then $f^{'} (x)$ at $x = e$ is $\dots\dots\dots$

JEE Advanced 1982

If $y = f (\frac{2 x - 1}{x ^{2} + 1})$ and $f^{'} (x) = sin x^{2}$ , then $\frac{d y}{d x} = \dots\dots\dots$