Answer
$$\ y^{\prime} =2\left(x+x^{2 x}+x^{2 x} \ln x\right)$$
Work Step by Step
Given:
$y=x^{2}+x^{2 x}$
We get:
\begin{aligned} &y=x^{2}+x^{2 x} \Rightarrow y-x^{2}=x^{2 x} \\
&\Rightarrow \ln \left(y-x^{2}\right)=\ln x^{2 x} \\
&\Rightarrow \ln \left(y-x^{2}\right)
=2 x \ln x \\
&\text{ differentiate with respect to } \ x, \\
&\Rightarrow \frac{1}{y-x^{2}}\left(y^{\prime}-2 x\right)=2 x \cdot \frac{1}{x}+2 \cdot \ln x=2+2 \ln x\\
&\Rightarrow y^{\prime}-2 x=\left(y-x^{2}\right)(2+2 \ln x) \\
&\Rightarrow y^{\prime}
=\left(\left(x^{2}+x^{2 x}\right)-x^{2}\right)(2+2 \ln x)+2 x\\
&\ \ \ \ \ \ \ \ \ \ =2\left(x+x^{2 x}+x^{2 x} \ln x\right)\end{aligned}
