# Find the values

Question:

Find $\frac{\mathrm{dy}}{\mathrm{dx}}$, when

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

Solution:

Here,

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

$=e^{\log x^{x}}+e^{\log x^{\frac{1}{x}}}$

$y=e^{x \log x}+e^{\left(\frac{1}{x} \log x\right)}$

[ Sincelog $a^{b}=b \log a$ ]

Differentiating it with respect to $x$ using the chain rule and product rule,

$\frac{d y}{d x}=\frac{d}{d x}\left(e^{x \log x}\right)+\frac{d}{d x}\left(e^{\frac{1}{x} \log x}\right)$

$=e^{x \log x}+\frac{d}{d x}(x \log x)+e^{\frac{1}{x} \log x} \frac{d}{d x}\left(\frac{1}{x} \log x\right)$

$=e^{\log x^{x}}\left[x \frac{d}{d x}(\log x)+\log x \frac{d}{d x}(x)\right]+e^{\log x \frac{1}{x}}\left[\frac{1}{x} \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(\frac{1}{x}\right)\right]$

$=x^{x}\left[x\left(\frac{1}{x}\right)+\log x(1)\right]+x^{\frac{1}{x}}\left[\left(\frac{1}{x}\right)\left(\frac{1}{x}\right)+\log x\left(-\frac{1}{x^{2}}\right)\right]$

$=x^{x}[1+\log x]+x \frac{1}{x}\left(\frac{1}{x^{2}}-\frac{1}{x^{2}} \log x\right)$

$\frac{d y}{d x}=x^{x}[1+\log x]+x^{\frac{1}{x}} \frac{(1-\log x)}{x^{2}}$