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Let $f(x)>0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x)=\frac{1}{1+\log x} \cdot$ Then, $f^{\prime}(x)$ will be
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Verified Answer
The correct answer is:
$1+\log (f(x))$
Given,
$$
h\{f(x)\}=x
$$
differentiating w.r.t. $x$, we get
$$
\begin{array}{r}
h^{\prime}\{f(x)\}, f^{\prime}(x)=1 \\
f^{\prime}(x)=\frac{1}{h^{\prime}\{f(x)\}}
\end{array}
$$
$\begin{aligned} \Rightarrow \quad f^{\prime}(x)=\frac{1}{\frac{1}{1+\log \{f(x)\}}} & \\ \Rightarrow \quad\left[\because h^{\prime}(x)=\frac{1}{1+\log x}\right] \\ \Rightarrow & f^{\prime}(x)=1+\log (f(x)) \end{aligned}$
$$
h\{f(x)\}=x
$$
differentiating w.r.t. $x$, we get
$$
\begin{array}{r}
h^{\prime}\{f(x)\}, f^{\prime}(x)=1 \\
f^{\prime}(x)=\frac{1}{h^{\prime}\{f(x)\}}
\end{array}
$$
$\begin{aligned} \Rightarrow \quad f^{\prime}(x)=\frac{1}{\frac{1}{1+\log \{f(x)\}}} & \\ \Rightarrow \quad\left[\because h^{\prime}(x)=\frac{1}{1+\log x}\right] \\ \Rightarrow & f^{\prime}(x)=1+\log (f(x)) \end{aligned}$
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