If x = f ( t ) and y = g ( t ) are differentiable functions of t then d 2 y d x 2 is

f ′ ( t ) ⋅ g ″ ( t ) − g ′ ( t ) ⋅ f ″ ( t ) [ f ′ ( t ) ] 3

f ′ ( t ) ⋅ g ″ ( t ) − g ′ ( t ) ⋅ f ″ ( t ) [ f ′ ( t ) ] 2

g ′ ( t ) ⋅ f ″ ( t ) − f ′ ( t ) ⋅ g ″ ( t ) [ f ′ ( t ) ] 3

g ′ ( t ) ⋅ f ″ ( t ) + f ′ ( t ) ⋅ g ″ ( t ) [ f ′ ( t ) ] 3

EASY

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A differentiable function $f$ defined for all $x > 0$ satisfies $f (x^{2}) = x^{3}$ for all $x > 0$ . What is $f' (4) ?$

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Important Questions on Continuity and Differentiability

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

f (x) = x^{6} + 6^{x},

then

f^{'} (x)

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

x = \sqrt{2^{{cosec}^{- 1} t}}

and

y = \sqrt{2^{\sec^{- 1} t}}, (|t| \geq 1),

then

\frac{d y}{d x}

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

x = a (t - \frac{1}{t}), y = a (t + \frac{1}{t})

where

t

be the parameter then

\frac{d y}{d x} = ?

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

Let

p (x)

be a polynomial such that

p (x) - p^{'} (x) = x^{n}

where

n

is a positive integer. Then,

p (0)

equals

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

If the function

f (x)

defined by

f (x) = \frac{x^{100}}{100} + \frac{x^{99}}{99} + . . . . . . + \frac{x^{2}}{2} + x + 1,

then

f^{'} (0) =

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

Let

f

be a polynomial function such that

f (3 x) = f^{'} (x) . f^{''} (x),

for all

x \in R .

Then :

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

x = f (t)

and

y = g (t)

are differentiable functions of

t

then

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

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

The derivative of

{t a n}^{- 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

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

y = \sin^{- 1} \frac{1}{2} (\sqrt{1 + x} + \sqrt{1 - x})

then

y^{'}

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

x^{m} y^{n} = {(x + y)}^{m + n}

then

y^{'}

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

g

is the inverse of a function

f

and

f^{'} (x) = \frac{1}{1 + x^{5}},

then

g^{'} (x)

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

y = \frac{x + c}{1 + x^{2}}

where

c

is a constant, when

y

is stationary,

x y

is equal to -

HARD

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

f (x) = |\begin{matrix} \cos 2 x & \cos 2 x & \sin 2 x \\ - \cos x & \cos x & - \sin x \\ \sin x & \sin x & \cos x \end{matrix}|,

then

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

y = \frac{\sin^{2} x}{1 + \cot x} + \frac{\cos^{2} x}{1 + \tan x}

,

then

y^{'} (x)

is equal to

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

f (1) = 1, f^{'} (1) = 3

, then the derivative of

f (f (f (x))) + {(f (x))}^{2}

x = 1

is:

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

If then

x^{y} = e^{x - y},

then

\frac{d y}{d x}

x = 1

is ….

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

e^{y} + x y = e,

the ordered pair

(\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}})

x = 0

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

y = x^{x^{x^{. . . . . \infty}}},

then

y^{'}

is equal to

MEDIUM

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

x = 3 t a n t

and

y = 3 s e c t,

then the value of

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

t = \frac{π}{4},

is:

EASY

Mathematics>Differential Calculus>Continuity and Differentiability>Methods of Differentiation

Let

f (- x) = f (x)

. Then

f^{'} (x)

must be

A differentiable function f defined for all x>0 satisfies fx2=x3 for all x>0. What is f'4?

Important Questions on Continuity and Differentiability

Learn and Prepare for any exam you want !

A differentiable function $f$ defined for all $x > 0$ satisfies $f (x^{2}) = x^{3}$ for all $x > 0$ . What is $f' (4) ?$