Question

Differentiate the given functions w.r.t. x.

$(logx{)}^{cosx}$

Answer 1

Ley y = $(logx{)}^{cosx}$

Taking log on both sides, we get

$logy=log\left\{\right(logx{)}^{cosx}\}$

$\Rightarrow logy=cosx\times log\left(logx\right)$ [$\because log{m}^n=nlogm$]

Differentiating both sides w.r.t. x, we get

$\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}\left\{cosxlog\right(logx\left)\right\}$
$\Rightarrow \frac{1}{y}\frac{dy}{dx}=cosx\frac{1}{logx}.\frac{1}{x}+log\left(logx\right)(-sinx)$

$\left(Usingproductrule\frac{d}{dx}\right(u.v)=u\frac{dv}{dx}+v\frac{du}{dx})$
$\Rightarrow \frac{dy}{dx}=y\{\frac{cosx}{xlogx}-sinxlog(logx\left)\right\}$
$=(logx{)}^{cosx}\{\frac{cosx}{xlogx}-sinxlog(logx\left)\right\}$

Note: In a log function, if base is not given then, it is consider as e, then $\frac{d}{dx}\left(logx\right)=\frac{1}{x}$. If base is other than e. Then, $\frac{d}{dx}\left(lo{g}_{a}x\right)=lo{g}_{a}e.\frac{1}{x}$