Let g(x) be the inverse of the function f(x) | vedclass

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(C) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$f'(g(x)) \cd

Vedclass · Accepted Answer

$1 + (g(x))^3$

Vedclass · Answer

(C) Given that $g(x)$ is the inverse of $f(x)$,we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$f'(g(x)) \cdot g'(x) = 1$Therefore,$g'(x) = \frac{1}{f'(g(x))}$.Given $f'(x) = \frac{1}{1 + x^3}$,we substitute $g(x)$ for $x$ in the derivative expression:$f'(g(x)) = \frac{1}{1 + (g(x))^3}$.Substituting this back into the expression for $g'(x)$:$g'(x) = \frac{1}{\frac{1}{1 + (g(x))^3}} = 1 + (g(x))^3$.Thus,the correct option is $C$.

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to

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