Differentiate (sex)² in detail explaining every step

Let $y = (x e^x)^2$.
We can also write this as $y = x^2 (e^x)^2 = x^2 e^{2x}$.
Let's use the form $y = (x e^x)^2$ and apply the chain rule first.

The chain rule states that if $y = [f(x)]^n$, then $\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)$.
Here, $f(x) = x e^x$ and $n=2$.

The product rule states that if $g(x) = u(x)v(x)$, then $g'(x) = u'(x)v(x) + u(x)v'(x)$.
Here, $u(x) = x$ and $v(x) = e^x$.
So, $u'(x) = \frac{d}{dx}(x) = 1$.
And $v'(x) = \frac{d}{dx}(e^x) = e^x$.
Therefore,

The final answer is $2x e^{2x}(1+x)$.