Answer
$$\frac{{dy}}{{dx}} = x{e^{4x}}$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{4}x{e^{4x}} - \frac{1}{{16}}{e^{4x}} \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{4}x{e^{4x}} - \frac{1}{{16}}{e^{4x}}} \right] \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\frac{1}{4}x{e^{4x}}} \right] - \frac{d}{{dx}}\left[ {\frac{1}{{16}}{e^{4x}}} \right] \cr
& {\text{use the product rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{4}x\frac{d}{{dx}}\left[ {{e^{4x}}} \right] + \frac{1}{4}{e^{4x}}\frac{d}{{dx}}\left[ x \right] - \frac{1}{{16}}\frac{d}{{dx}}\left[ {{e^{4x}}} \right] \cr
& {\text{we can use the formula }}\frac{d}{{dx}}{e^u} = {e^u}\frac{{du}}{{dx}}\cr
&{\text{where }}u{\text{ is any differentiable function of }}x \cr
& \frac{{dy}}{{dx}} = \frac{1}{4}x\left( {4{e^{4x}}} \right) + \frac{1}{4}{e^{4x}}\left( 1 \right) - \frac{1}{{16}}\left( {4{e^{4x}}} \right) \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = x{e^{4x}} + \frac{1}{4}{e^{4x}} - \frac{1}{4}{e^{4x}} \cr
& \frac{{dy}}{{dx}} = x{e^{4x}} \cr} $$
