Find \(\frac{dy}{dx}\) if \(y=(\log x)^x+x^5\)

Answer: \(\frac{dy}{dx}=(\log x)^x\left[\log(\log x)+\frac{1}{\log x}\right]+5x^4\)

Let \(u=(\log x)^x\) and \(v=x^5\), so \(y=u+v\) and \(\frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}\).

\(\log u=x\log(\log x)\). Differentiating w.r.t. \(x\):
\(\frac{1}{u}\frac{du}{dx}=x\cdot\frac{1}{\log x}\cdot\frac{1}{x}+\log(\log x)\cdot 1=\frac{1}{\log x}+\log(\log x)\)
\(\therefore\frac{du}{dx}=(\log x)^x\left[\log(\log x)+\frac{1}{\log x}\right]\)

\(\frac{dv}{dx}=5x^4\).

\(\therefore\frac{dy}{dx}=(\log x)^x\left[\log(\log x)+\frac{1}{\log x}\right]+5x^4\).