If g is the inverse of f and f^{prime}(x)=fra | vedclass

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(A) Given that $g$ is the inverse of $f$,so $g(x) = f^{-1}(x)$.This implies $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the cha

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$1+[g(x)]^{2}$

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(A) Given that $g$ is the inverse of $f$,so $g(x) = f^{-1}(x)$.This implies $f(g(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule,we get:$f^{\prime}(g(x)) \cdot g^{\prime}(x) = 1$.Therefore,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$.Given $f^{\prime}(x) = \frac{1}{1+x^{2}}$,we substitute $g(x)$ for $x$:$f^{\prime}(g(x)) = \frac{1}{1+[g(x)]^{2}}$.Substituting this into the expression for $g^{\prime}(x)$:$g^{\prime}(x) = \frac{1}{\frac{1}{1+[g(x)]^{2}}} = 1 + [g(x)]^{2}$.

If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^{2}}$,then $g^{\prime}(x)$ is equal to

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