JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

\mathrm{g}

is the inverse of

\mathrm{f}

and

\mathrm{f}^{\prime}(x)=\frac{1}{1+x^3}

, then find

g^{\prime}(x)

View answer and explanation

Correct option: C

1+[g(x)]^3

Option C (

1+[g(x)]^3

) is correct because by differentiating

f[g(x)] = x

with respect to x and using the given

f'(x)

, we find that

g'(x) = 1 + [g(x)]^3

More Questions from DIFFERENTIATION

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

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

x

y

x>0

x y=1

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

[x]

x

x=k, k

f(x)

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],

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

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

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

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