Question

What is derivative of $[\frac{1}{x}{]}^x$

Answer 1

Consider the given function

$y={\left(\frac{1}{x}\right)}^x$

Taking log both side and we get,

$\log y=\log {\left(\frac{1}{x}\right)}^x$

$\log y=x\log \left(\frac{1}{x}\right)$

On differentiation this equation with respect to x and we get,

$\frac{d}{dx}\log y=\frac{d}{dx}(x.\log \left(\frac{1}{x}\right))$

$\frac{d}{dx}\log y=x.\frac{d}{dx}\log \left(\frac{1}{x}\right)+\log \left(\frac{1}{x}\right)\frac{d}{dx}x$

$\frac{1}{y}\frac{dy}{dx}=x.\frac{d}{dx}{\left(\mathrm{logx}\right)}^{-1}+\log \left(\frac{1}{x}\right)\frac{d}{dx}x$

$\frac{1}{y}\frac{dy}{dx}=-x\frac{d}{dx}\log x+\log \left(\frac{1}{x}\right)\frac{d}{dx}x$

$\frac{1}{y}\frac{dy}{dx}=-x\frac{1}{x}+\log \left(\frac{1}{x}\right)$

$\frac{dy}{dx}=y(-1+\log \left(\frac{1}{x}\right))$

$\frac{dy}{dx}=y(\log \left(\frac{1}{x}\right)-1)$

$\frac{dy}{dx}={\left(\frac{1}{x}\right)}^x(\log \left(\frac{1}{x}\right)-1)$

Hence, this is the answer.