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

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(B) Given that $g(x)$ is the inverse of the function $f(x)$,we have $g(x) = f^{-1}(x)$.This implies $f(g(x)) = x$.Differentiating both sides with resp

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

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(B) Given that $g(x)$ is the inverse of the function $f(x)$,we have $g(x) = f^{-1}(x)$.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,$g^{\prime}(x) = \frac{1}{f^{\prime}(g(x))}$ ... $(i)$Given $f^{\prime}(x) = \frac{1}{1+x^3}$,we substitute $x$ with $g(x)$ to get:$f^{\prime}(g(x)) = \frac{1}{1+(g(x))^3}$ ... $(ii)$Substituting $(ii)$ into $(i)$,we get:$g^{\prime}(x) = \frac{1}{\frac{1}{1+(g(x))^3}} = 1+(g(x))^3$.

If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^3}$,then $g^{\prime}(x)$ is

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