How do you differentiate f(x)=e^(4x)sinx using the product rule?

Dec 16, 2016

$f ' \left(x\right) = {e}^{4 x} \cos x + 4 {e}^{4 x} \sin x$

Explanation:

If you are studying maths, then you should learn the Product Rule for Differentiation, and practice how to use it:

$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$, or, $\left(u v\right) ' = \left(\mathrm{du}\right) v + u \left(\mathrm{dv}\right)$

I was taught to remember the rule in words; "The first times the derivative of the second plus the derivative of the first times the second ".

This can be extended to three products:

$\frac{d}{\mathrm{dx}} \left(u v w\right) = u v \frac{\mathrm{dw}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}} w + \frac{\mathrm{du}}{\mathrm{dx}} v w$

So with $f \left(x\right) = {x}^{2} \sin x \tan x$ we have;

$\left\{\begin{matrix}\text{Let "u = e^(4x) & => & (du)/dx = 4e^(4x)' \\ "And } v = \sin x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = \cos x '\end{matrix}\right.$

Applying the product rule we get:

$\text{ } \frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + \frac{\mathrm{du}}{\mathrm{dx}} v$
$\frac{d}{\mathrm{dx}} \left({e}^{4 x} \sin x\right) = \left({e}^{4 x}\right) \left(\cos x\right) + \left(4 {e}^{4 x}\right) \left(\sin x\right)$
$\frac{d}{\mathrm{dx}} \left({e}^{4 x} \sin x\right) = {e}^{4 x} \cos x + 4 {e}^{4 x} \sin x$