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

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(B) Since $g$ is the inverse of $f$,we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule,we get $f'(g(x)) \cdot g

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$\frac{1 + a^2}{a^{10}}$

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(B) Since $g$ is the inverse of $f$,we have $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule,we get $f'(g(x)) \cdot g'(x) = 1$.Substituting $x = 2$,we get $f'(g(2)) \cdot g'(2) = 1$.Given $g(2) = a$,this becomes $f'(a) \cdot g'(2) = 1$,which implies $g'(2) = \frac{1}{f'(a)}$.We are given $f'(x) = \frac{x^{10}}{1 + x^2}$,so $f'(a) = \frac{a^{10}}{1 + a^2}$.Therefore,$g'(2) = \frac{1}{\frac{a^{10}}{1 + a^2}} = \frac{1 + a^2}{a^{10}}$.

Let $g$ be the inverse function of $f$ and $f'(x) = \frac{x^{10}}{1 + x^2}$. If $g(2) = a$,then $g'(2)$ is equal to:

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