Given that $y = \frac{e^{2x}}{\sqrt{1-4x}}$, show that $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4e^{2x}(1-2x)}{(1-4x)\sqrt{1-4x}}$.

1. Differentiate the numerator and the denominator

Let $u = e^{2x}$ and $v = (1-4x)^{\frac{1}{2}}$.
$\frac{\mathrm{d}u}{\mathrm{d}x} = 2e^{2x}$
$\frac{\mathrm{d}v}{\mathrm{d}x} = \frac{1}{2}(1-4x)^{-\frac{1}{2}}(-4) = -2(1-4x)^{-\frac{1}{2}}$

2. Apply the quotient rule

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(1-4x)^{\frac{1}{2}}(2e^{2x}) - e^{2x}(-2(1-4x)^{-\frac{1}{2}})}{(\sqrt{1-4x})^2}$

3. Simplify the expression to match the required form

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{2x}(1-4x)^{\frac{1}{2}} + 2e^{2x}(1-4x)^{-\frac{1}{2}}}{1-4x}$
Factor out $2e^{2x}(1-4x)^{-\frac{1}{2}}$ from the numerator:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{2x}(1-4x)^{-\frac{1}{2}} [ (1-4x) + 1 ]}{1-4x}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2e^{2x}(2-4x)}{(1-4x)(1-4x)^{\frac{1}{2}}}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4e^{2x}(1-2x)}{(1-4x)\sqrt{1-4x}}$ (shown)