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

Feb 2, 2018

$\frac{- 2 \sin \left(2 x\right) \cdot \left({e}^{2 x} - x\right) - \left(2 {e}^{2 x} - 1\right) \cdot \cos \left(2 x\right)}{{e}^{2 x} - x} ^ 2$

#### Explanation:

The quotient rule states that, given $\left(\frac{f}{g}\right) '$, (mind that this is in Newton's notation), the derivative is $\frac{f ' g - f g '}{g} ^ 2$

Here, $f = \cos 2 x$, and $g = {e}^{2 x} - x$.

Let's tackle the problem a step at a time.

First, $f '$.

The derivative of $\cos 2 x$ cannot be calculated directly. We must use the chain rule, which states that $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

Here, $f = \cos u$, and $u = 2 x$

The derivative of $\cos u = - \sin u$, and the derivative of $2 x = 2$.

Multiplying: $- 2 \sin u$

$f ' = - 2 \sin 2 x$

Next, $g '$.

Again, the derivative of ${e}^{2} x - x$ cannot be done directly. The difference rule states that $\left(f - g\right) ' = f ' - g '$.

So first, $\frac{d}{\mathrm{dx}} {e}^{2 x}$. Use the chain rule again, here $f = {e}^{u}$ and $g = 2 x$.

The derivative of ${e}^{u} = {e}^{u}$, and of $2 x = 2$

So this becomes $2 {e}^{u}$.

$g ' = 2 {e}^{2 x} - 1$. The $1$ is the derivative of $x$.

Now we can simply input.

$\frac{f ' g - f g '}{g} ^ 2$

$\frac{\left(- 2 \sin 2 x\right) \left({e}^{2 x} - x\right) - \left(\cos 2 x\right) \left(2 {e}^{2 x} - 1\right)}{{e}^{2 x} - x} ^ 2$

$\frac{- 2 \sin \left(2 x\right) \cdot \left({e}^{2 x} - x\right) - \left(2 {e}^{2 x} - 1\right) \cdot \cos \left(2 x\right)}{{e}^{2 x} - x} ^ 2$