Question

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

• A. $1+\log \left(f\right(x\left)\right)$
• B. $1+f\left(x\right)$
• C. $1-\log \left(f\right(x\left)\right)$
• D. $\log f\left(x\right)$

Answer 1

The correct option is A $1+\log \left(f\right(x\left)\right)$

$f$ is the inverse function of $h$
$\Rightarrow h\left(f\right(x\left)\right)=x$
Differentiating w.r.t. $x$ ,
$h^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=1$

$\Rightarrow \frac{1}{1+\log f\left(x\right)}{f}^{\prime }\left(x\right)=1$
$\Rightarrow f^{\prime }\left(x\right)=1+\log f\left(x\right)$