Let g(x) be the inverse of an invertible func | vedclass

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(B) Since $g(x)$ is the inverse of the function $f(x)$,we have $g(f(x)) = x$ for all $x$ in the domain.Differentiating both sides with respect to $x$

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$\frac{1}{f'(c)}$

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(B) Since $g(x)$ is the inverse of the function $f(x)$,we have $g(f(x)) = x$ for all $x$ in the domain.Differentiating both sides with respect to $x$ using the chain rule:$\frac{d}{dx}[g(f(x))] = \frac{d}{dx}(x)$$g'(f(x)) \cdot f'(x) = 1$Assuming $f'(x) 
eq 0$,we can write:$g'(f(x)) = \frac{1}{f'(x)}$Substituting $x = c$ into the equation:$g'(f(c)) = \frac{1}{f'(c)}$Thus,the correct option is $B$.

Let $g(x)$ be the inverse of an invertible function $f(x)$ which is differentiable at $x = c$. Then $g'(f(c))$ equals:

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