Let f(x) 0 for all x and f^{prime}(x) exist | vedclass

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(A) Given that $f$ is the inverse function of $h$, we have $h(f(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$h^{\prim

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$1 + \log (f(x))$

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(A) Given that $f$ is the inverse function of $h$, we have $h(f(x)) = x$.Differentiating both sides with respect to $x$ using the chain rule:$h^{\prime}(f(x)) \cdot f^{\prime}(x) = 1$Therefore, $f^{\prime}(x) = \frac{1}{h^{\prime}(f(x))}$.Given $h^{\prime}(x) = \frac{1}{1 + \log x}$, we substitute $f(x)$ for $x$ to get $h^{\prime}(f(x)) = \frac{1}{1 + \log(f(x))}$.Substituting this into the expression for $f^{\prime}(x)$:$f^{\prime}(x) = \frac{1}{\frac{1}{1 + \log(f(x))}} = 1 + \log(f(x))$.

Let $f(x) > 0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x) = \frac{1}{1 + \log x}$, then $f^{\prime}(x)$ will be

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