Question

Find $\frac{dy}{dx}$

$y=x^x+x^{1/x}$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have},y=x^x+x^{\frac{1}{x}} \\ \Rightarrow y=e^{\log x^x}+e^{\log x^{\frac{1}{x}}} \\ \Rightarrow y=e^{x\log x}+e^{\left(\frac{1}{x}\log x\right)} \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(e^{x\log x}\right)+\frac{d}{dx}\left(e^{\frac{1}{x}\log x}\right) \\ =e^{x\log x}\frac{d}{dx}\left(x\log x\right)+e^{\frac{1}{x}\log x}\frac{d}{dx}\left(\frac{1}{x}\log x\right) \\ =e^{\log x^x}\left[x\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(x\right)\right]+e^{\log x^{\frac{1}{x}}}\left[\frac{1}{x}\frac{d}{dx}\left(\log x\right)+\log x\frac{d}{dx}\left(\frac{1}{x}\right)\right] \\ ={\left(x\right)}^x\left[x\left(\frac{1}{x}\right)+\log x\left(1\right)\right]+x^{\frac{1}{x}}\left[\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)+\log x\left(-\frac{1}{x^2}\right)\right] \\ ={\left(x\right)}^x\left[1+\log x\right]+x^{\frac{1}{x}}\left(\frac{1}{x^2}-\frac{1}{x^2}\log x\right) \\ =x^x\left[1+\log x\right]+x^{\frac{1}{x}}\left(\frac{1-\log x}{x^2}\right) \end{gathered}$$