JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

\mathrm{G}(\mathrm{x})=\frac{1}{\sqrt{25-\mathrm{x}^2}}

, then

x_{x \rightarrow 0} \frac{\mathrm{G}(\mathrm{x}+\mathrm{h})-\mathrm{G}(4)}{\mathrm{x}-4}

has the value

View answer and explanation

Correct option: D

none of these

Option D (none of these) is correct because the limit represents the derivative of G(x) at x=4, and the calculated derivative does not match any of the given options.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

f(x)=|x|^{|\tan x|}

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

then

\frac{{dy}}{{dx}}

is equal to

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}