Question

$y=(\cos x{)}^{\log x}+(\log x{)}^x$, then $\frac{dy}{dx}=(\cos x{)}^{\log x}\left[\log x\right(-\tan x)+\frac{1}{x}\log \cos x]+(\log x{)}^x[\frac{1}{\log x}+\log (\log x\left)\right]$, if true enter 1 else enter 0.

Answer 1

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$
$u=(\cos x{)}^{\log x}$
taking log both side ,
$\log u=\log x\left(\log \cos x\right)$
Differentiating both side w.r.t $x$
$\frac{1}{u}\frac{du}{dx}=\frac{\log \cos x}{x}-\frac{\log x.\left(\sin x\right)}{\cos x}$
$\frac{du}{dx}=(\cos x{)}^{\log x}\left[\log x\right(-\tan x)+\frac{\log \cos x}{x}]$
Now $v=(\log x{)}^x\Rightarrow \log v=x\log (\log x)$
$\Rightarrow \frac{1}{v}\frac{dv}{dx}=\log \left(\log x\right)+\frac{1}{\log x}$
$\frac{dv}{dx}=(\log x{)}^x[\frac{1}{\log x}+\log (\log x\left)\right]$
$\therefore \frac{dy}{dx}=(\cos x{)}^{\log x}\left[\log x\right(-\tan x)+\frac{\log \cos x}{x}]+(\log x{)}^x[\frac{1}{\log x}+\log (\log x\left)\right]$