Question

Differentiate the following functions w.r.t.$x$
$a^{\log (1+\log x)}$

Answer 1

Let $v=1+\log x,u=\log v$ and $y=a^u$

then $v=1+\log x\Rightarrow \frac{dv}{dx}=0+\frac{1}{x}=\frac{1}{x}$ since $\frac{d}{dx}\left(\log x\right)=\frac{1}{x}$

$u=\log u\Rightarrow \frac{du}{dv}=\frac{d}{du}\left(\log u\right)=\frac{1}{u}$

and $y=a^u\Rightarrow \frac{dy}{du}=a^u\log a$ since $\frac{d}{dx}{a}^x=a^x\log a$

Using chain rule we have $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dv}\times \frac{dv}{dx}$ by substituting the above values

$=a^u\log a\times \frac{1}{u}\times \frac{1}{x}$

$=a^{\log v}\log a\times \frac{1}{\log v}\times \frac{1}{x}$

$=a^{\log (1+\log x)}\log a\times \frac{1}{\log (1+\log x)}\times \frac{1}{x}$

$=\frac{a^{\log (1+\log x)}\log a}{x(1+\log x)}$