How do you differentiate g(x) = (x/2)e^(2x) using the product rule?

Sep 14, 2016

$\frac{\mathrm{dg}}{\mathrm{dx}} = {e}^{2 x} / 2 + x {e}^{2 x}$

Explanation:

Product rule states if $g \left(x\right) = f \left(x\right) h \left(x\right)$

then $\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dx}} \times h \left(x\right) + \frac{\mathrm{dh}}{\mathrm{dx}} \times f \left(x\right)$

Hence as $g \left(x\right) = \left(\frac{x}{2}\right) {e}^{2 x}$

$\frac{\mathrm{dg}}{\mathrm{dx}} = \frac{1}{2} \times {e}^{2 x} + 2 \times {e}^{2 x} \times \frac{x}{2}$

Here we have also used the concept of function of a function and used chain rule for it. As we have differentiated ${e}^{2 x}$ w.r.t. $2 x$, we must multiply by differential of $2 x$ w.r.t $x$ i.e. by $2$.

So $\frac{\mathrm{dg}}{\mathrm{dx}} = {e}^{2 x} / 2 + x {e}^{2 x}$