For x > 1 , if 2 x 2 y = 4 e 2 x - 2 y , then 1 + log e ⁡ 2 x 2 d y d x is equal to

x log e ⁡ 2 x - log e ⁡ 2 x

x log e ⁡ 2 x + log e ⁡ 2 x

HARD

Earn 100

Let D be the domain of a twice differentiable function f. For all $x \in D, f^{''} (x) + f (x) = 0$

And $f (x) = \int g (x) d x +$ constant. If $h (x) = {(f (x))}^{2}$ and $h (0) = 5$ then $h (2015) - h (2014) =$

14.29% studentsanswered this correctly

Important Questions on Limits and Derivatives

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

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

,

then

y^{'} (x)

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

g (x)

is the inverse function of

f (x)

and

f^{'} (x) = \frac{1}{1 + x^{4}},

then

g^{'} (x)

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

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

then

f^{'} (x)

is equal to

HARD

Mathematics>Differential Calculus>Limits and Derivatives>First Principle 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

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

y = \sec (\tan^{- 1} x)

, then

\frac{d y}{d x}

x = 1

is equal to

HARD

Mathematics>Differential Calculus>Limits and Derivatives>First Principle 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>Limits and Derivatives>First Principle of Differentiation

Let

f

and

g

be differentiable functions such that

f (3) = 5, g (3) = 7, f^{'} (3) = 13, g^{'} (3) = 6, f^{'} (7) = 2

and

g^{'} (7) = 0

.

h (x) = (f o g) (x),

then

h^{'} (3) =

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

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

EASY

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

\sin y = x \sin (a + y)

, then

\frac{d y}{d x}

EASY

Mathematics>Differential Calculus>Limits and Derivatives>First Principle 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) =

EASY

Mathematics>Differential Calculus>Limits and Derivatives>First Principle 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>Limits and Derivatives>First Principle of Differentiation

For

x > 1

, if

{(2 x)}^{2 y} = 4 e^{2 x - 2 y}

, then

{(1 + \log_{e} 2 x)}^{2} \frac{d y}{d x}

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

\int e^{\sec x} (\sec x \tan x f (x) + (\sec x \tan x + \sec^{2} x)) d x = e^{\sec x} f (x) + C,

then a possible choice of

f (x)

is:

EASY

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

Let

f : f (- x) \to f (x)

be a differentiable function. If

f

is even, then

f^{'} (0)

is equal to

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

Let

f (x) = \log_{e} (s i n x), (0 < x < π)

and

g (x) = \sin^{- 1} (e^{- x}), (x \geq 0) .

α

is a positive real number such that

a = {(f o g)}^{'} (α)

and

b = (f o g) (α),

then

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

Let

y = y (x)

be a function of

x

satisfying

y \sqrt{1 - x^{2}} = k - x \sqrt{1 - y^{2}}

, where

k

is a constant and

y (\frac{1}{2}) = - \frac{1}{4}

. Then

\frac{d y}{d x}

x = \frac{1}{2}

, is equal to

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

y (α) = \sqrt{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}

α = \frac{5 π}{6}

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

If for

x \in (0, \frac{1}{4}),

the derivative of

\tan^{- 1} (\frac{6 x \sqrt{x}}{1 - 9 x^{3}})

\sqrt{x} \cdot g (x)

, then

g (x)

equals:

HARD

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

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 :

MEDIUM

Mathematics>Differential Calculus>Limits and Derivatives>First Principle of Differentiation

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

where

c

is a constant, when

y

is stationary,

x y

is equal to -

Let D be the domain of a twice differentiable function f. For all x∈D, f′′x+ fx= 0 And fx= ∫gxdx+ constant. If hx=fx2 and h0=5 then h2015-h2014=

Important Questions on Limits and Derivatives

Learn and Prepare for any exam you want !

Let D be the domain of a twice differentiable function f. For all $x \in D, f^{''} (x) + f (x) = 0$

And $f (x) = \int g (x) d x +$ constant. If $h (x) = {(f (x))}^{2}$ and $h (0) = 5$ then $h (2015) - h (2014) =$