Question

Let $y=(logx{)}^x+x^{x\cos x}$ , then find $\frac{dy}{dx}$.

Answer 1

$y={\left(\log x\right)}^x+x^{x\cos x}$

$\frac{dy}{dx}=\frac{d}{dx}\left({\left(\log x\right)}^x\right)+\frac{d}{dx}\left(x^{x\cos x}\right)$

$y_1=\log x^x$

$\log y_1=x\log \left(\log x\right)$

$\frac{1}{y_1}\frac{d{y}_1}{dx}=x.\frac{1}{\log x}\times \frac{1}{x}+\log \left(\log x\right)$

$\frac{d{y}_1}{dx}=\left(\frac{1}{\log x}9+\log \log x\right)\times (\log x{)}^2$

$y_2=x^{x\cos x}$

$\log y_2=x\cos x\log x$

$\frac{1}{y_2}\frac{d{y}_2}{dx}=x\cos x.\frac{1}{x}+x\log x(-\sin x)+\cos x\log x$

$\frac{d{y}_2}{dx}=x^{x\cos x}\left(\log x\right(\cos x-x\sin x)+\cos x)$

$\frac{dy}{dx}=(\log x{)}^x\left[\frac{1}{\log x}+\log \log x\right]+x^{x\cos x}\left[\log x(\cos x-x\sin x)+\cos x\right]$