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JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

Let

f(x)=-4 \cdot e^{\frac{2-x}{4}}+10+2 x+x^3+x^5

and

g(x)=f^{-1}(x)

, then the value of

g^{\prime}(50)

is

View answer and explanation

Correct option: C

\frac{1}{95}

Option C (

\frac{1}{95}

) is correct because using the inverse function derivative formula

g'(x) = \frac{1}{f'(g(x))}

, we find

g'(50) = \frac{1}{f'(2)} = \frac{1}{95}

.

More Questions from DIFFERENTIATION

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

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is equal to Let

x

y

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x y=1

. The minimum value of

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

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denotes the integer part of

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x=k, k

being an integer, the left derivative of

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),

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

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