Question

Find $\frac{dy}{dx}$ if $y={\left(x+\frac{1}{x}\right)}^x$

Answer 1

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

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

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

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

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

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

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

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