Question

If $y=x^x+\frac{1}{x^x}$, then $\frac{dy}{dx}=$

Answer 1

$y=x^x+\frac{1}{x^x}$

$x^4=p$

apply log:

$\log x+\frac{x1}{x}=\frac{1}{p}\frac{dp}{dx}$

$\frac{dp}{dx}=p(1+\log x)$

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

$y=p+\frac{1}{p}$

$\frac{dy}{dp}=(1-\frac{1}{p^2})$

$\Rightarrow \frac{dy}{dx}=(1-\frac{1}{p^2})\times x^x(1+\log x)$

$\Rightarrow \frac{dp}{dx}\left(\frac{(x^x{)}^2-1}{(x^x{)}^2}\right)x^x(1+\log x)=\left(\frac{x^{2x}-1}{x^x}\right)$

$\therefore \frac{dy}{dx}=\left(\frac{x^{2x}-1}{x^4}\right)(1+\log x)$