How do you differentiate f(x) = (cos2x)/(e^(2x)+4x) using the quotient rule?

Nov 6, 2016

$f ' \left(x\right) = \frac{- 2 \left({e}^{2 x} + 4 x\right) \left(\sin x\right) - \left(2 {e}^{2 x} + 4\right) \left(\cos 2 x\right)}{{e}^{2 x} + 4 x} ^ 2$

Explanation:

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

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$, or less formally, $\left(\frac{u}{v}\right) ' = \frac{v \left(\mathrm{du}\right) - u \left(\mathrm{dv}\right)}{v} ^ 2$

I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.

So with $f \left(x\right) = \frac{\cos 2 x}{{e}^{2} x + 4 x}$ Then

$\left\{\begin{matrix}\text{Let "u=cos2x & => & (du)/dx=-2sinx \\ "And } v = {e}^{2 x} + 4 x & \implies & \frac{\mathrm{dv}}{\mathrm{dx}} = 2 {e}^{2 x} + 4\end{matrix}\right.$

$\therefore \frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \frac{\mathrm{du}}{\mathrm{dx}} - u \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$
$\therefore f ' \left(x\right) = \frac{\left({e}^{2 x} + 4 x\right) \left(- 2 \sin x\right) - \left(\cos 2 x\right) \left(2 {e}^{2 x} + 4\right)}{{e}^{2 x} + 4 x} ^ 2$
$\therefore f ' \left(x\right) = \frac{- 2 \left({e}^{2 x} + 4 x\right) \left(\sin x\right) - \left(2 {e}^{2 x} + 4\right) \left(\cos 2 x\right)}{{e}^{2 x} + 4 x} ^ 2$