Question

Find $\frac{dy}{dx}$

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

Answer 1

$$\begin{gathered} \mathrm{Let}y=x^{\log x}+{\left(\log x\right)}^x \\ \mathrm{Also},\mathrm{let}u={\left(\log x\right)}^x\mathrm{and}v=x^{\log x} \\ \therefore y=v+u \\ \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}+\frac{du}{dx}...\left(i\right) \\ \mathrm{Now},u={\left(\log x\right)}^x \\ \Rightarrow \log u=\log \left[{\left(\log x\right)}^x\right] \\ \Rightarrow \log u=x\log \left(\log x\right) \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{u}\frac{du}{dx}=\log \left(\log x\right)\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left[\log \left(\log x\right)\right] \\ \Rightarrow \frac{du}{dx}=u\left[\mathrm{og}\left(\log x\right)+x\frac{1}{\log x}\frac{d}{dx}\left(\log x\right)\right] \\ \Rightarrow \frac{du}{dx}={\left(\log x\right)}^x\left[\log \left(\log x\right)+\frac{x}{\log x}\times \frac{1}{x}\right] \\ \Rightarrow \frac{du}{dx}={\left(\log x\right)}^x\left[\log \left(\log x\right)+\frac{1}{\log x}\right]...\left(ii\right) \\ \mathrm{Also},v=x^{\log x} \\ \Rightarrow \log v=\log x^{\log x} \\ \Rightarrow \log v=\log x\log x={\left(\log x\right)}^2 \end{gathered}$$
Differentiating both sides with respect to x,
$$\begin{gathered} \frac{1}{v}\frac{dv}{dx}=\frac{d}{dx}\left[{\left(\log x\right)}^2\right] \\ \Rightarrow \frac{1}{v}\frac{dv}{dx}=2\left(\log x\right)\frac{d}{dx}\left(\log x\right) \\ \Rightarrow \frac{dv}{dx}=2v\left(\log x\right)\frac{1}{x} \\ \Rightarrow \frac{dv}{dx}=2x^{\log x}\frac{\log x}{x} \\ \Rightarrow \frac{dv}{dx}=2x^{\log x}\frac{\log x}{x}...\left(iii\right) \\ \mathrm{From}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{obtain} \\ \frac{dy}{dx}=2x^{\log x}\frac{\log x}{x}+{\left(\log x\right)}^x\left[\log \left(\log x\right)+\frac{1}{\log x}\right] \end{gathered}$$