Question

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

Answer 1

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

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

$\Rightarrow logx=xlog(x+\frac{1}{x})$

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

$\Rightarrow \frac{1}{u}\frac{dy}{dx}=\frac{(x-\frac{1}{x})}{(x+\frac{1}{x})}+log(x+\frac{1}{x})$

$\Rightarrow \frac{dy}{dx}=u[\frac{(x-\frac{1}{x})}{(x+\frac{1}{x})}+log(x+\frac{1}{x})]$

Let $v=x^{x+\frac{1}{x}}$

$logv=(x+\frac{1}{x})logx$

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=(1-\frac{1}{x^2})logx+(x+\frac{1}{x})\frac{1}{x}$

$\Rightarrow \frac{dv}{dx}=v[(1-\frac{1}{x^2})logx+(1+\frac{1}{x^2})]$

$\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$

$=u[\frac{x-\frac{1}{x}}{x+\frac{1}{x}}+log(x+\frac{1}{x})]+v[(1-\frac{1}{x^2})logx+(1+\frac{1}{x^2})]$

${(x+\frac{1}{x^2})}^x[\frac{x-\frac{1}{x}}{x+\frac{1}{x}}+log(x+\frac{1}{x})]+x^{x+\frac{1}{x}}[(1-\frac{1}{x^2})logx+x^{x+\frac{1}{x}}(1+\frac{1}{x^2})]$.