Question

Differentiate ${\left(\begin{array}{c}x+\frac{1}{x}\end{array}\right)}^x$ with respect to $x$.

Answer 1

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

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

Differentiating

$\frac{y^{\prime }}{y}=\ln (x+\frac{1}{x})+\frac{x}{(x+\frac{1}{x})}\times (1-\frac{1}{x^2})$

$=\ln (x+\frac{1}{x})+\frac{(x-\frac{1}{x})}{(x+\frac{1}{x})}$

$=\ln (x+\frac{1}{x})+\frac{x^2-1}{x^2+1}$

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