Differentiate the following with respect to $x$.

1. Apply the chain rule to the outer power function

$\frac{d}{dx} \left[ 3 \tan^2 \left( \frac{x}{2} \right) \right] = 3 \cdot 2 \tan \left( \frac{x}{2} \right) \cdot \frac{d}{dx} \left[ \tan \left( \frac{x}{2} \right) \right]$

2. Differentiate the inner trigonometric function and simplify

$= 6 \tan \left( \frac{x}{2} \right) \cdot \sec^2 \left( \frac{x}{2} \right) \cdot \frac{1}{2} \\ = 3 \tan \left( \frac{x}{2} \right) \sec^2 \left( \frac{x}{2} \right)$

1. Identify the functions for the product rule and find their derivatives

Let $u = x^2$ and $v = e^{-x}$.
Then $\frac{du}{dx} = 2x$ and $\frac{dv}{dx} = -e^{-x}$.

2. Apply the product rule

$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$
$\frac{d}{dx}(e^{-x} x^2) = x^2(-e^{-x}) + e^{-x}(2x)$

3. Simplify the expression by factorising

$= -x^2 e^{-x} + 2x e^{-x}$
$= x e^{-x}(2 - x)$