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

If y = ( cos x ) 2 x , then d y d x is equal to

2 ( cos x ) 2 x [ log ( cos x ) - x tan x ]

2 ( cos x ) 2 x ( sin x - x tan x )

2 ( cos x ) 2 x [ log ( cos x ) + x tan x ]

2 ( sin x ) 2 x [ log ( cos x ) - x tan x ]

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If $y = x + \frac{1}{x + \frac{1}{x + \frac{1}{x +} . . . \infty}}$ , then the differential coefficient $\frac{d y}{d x}$ equals:

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Important Questions on Differential Coefficient

EASY

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

If then

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

then

\frac{d y}{d x}

x = 1

is ….

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>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

EASY

Mathematics>Differential Calculus>Differential Coefficient>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>Differential Coefficient>Methods of Differentiation

\sqrt[3]{y} \sqrt{x} = \sqrt[6]{{(x + y)}^{5}},

then

\frac{d y}{d x} =

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

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

then

y^{'}

is equal to

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

x = 2 \sin θ - \sin 2 θ

and

y = 2 \cos θ - \cos 2 θ,

θ \in [0, 2 π],

then

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

θ = π

is:

HARD

Mathematics>Differential Calculus>Differential Coefficient>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

HARD

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

The derivative of

\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})

with respect to

\tan^{- 1} (\frac{2 x \sqrt{1 - x^{2}}}{1 - 2 x^{2}})

x = \frac{1}{2}

is :

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>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>Differential Coefficient>Methods of Differentiation

x = f (t)

and

y = g (t)

are differentiable functions of

t

then

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

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

x^{3} + 2 x y + \frac{1}{3} y^{3} = \frac{11}{3}

, then

\frac{d y}{d x}

(2, - 1)

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>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:

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

y = (\cos x)^{2 x}

, then

\frac{d y}{d x}

is equal to

EASY

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

x = s i n θ, y = {s i n}^{3} θ,

then

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

θ

\frac{π}{2}

is ….

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

Suppose a differentiable function

f (x)

satisfies the identity

f (x + y) = f (x) + f (y) + x y^{2} + x^{2} y,

for all real

x

and

y

. If

\lim_{x \to 0} \frac{f (x)}{x} = 1,

then

f^{'} (3)

is equal to :

EASY

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

The derivative of

\cos^{- 1} (2 x^{2} - 1)

w.r.t.,

\cos^{- 1} x

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

x^{m} y^{n} = 2 {(x + y)}^{m + n}

, the value of

\frac{d y}{d x}

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>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} = ?

MEDIUM

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

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

then

y^{'}

is equal to

EASY

Mathematics>Differential Calculus>Differential Coefficient>Methods of Differentiation

Let

f (- x) = f (x)

. Then

f^{'} (x)

must be

If y=x+1x+1x+1x+...∞, then the differential coefficient dydx equals:

Important Questions on Differential Coefficient

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If $y = x + \frac{1}{x + \frac{1}{x + \frac{1}{x +} . . . \infty}}$ , then the differential coefficient $\frac{d y}{d x}$ equals: