# How do you use the Quotient Rule to find the derivative of f(x) = [1 + sin(2x)]/[1 - sin(2x)]?

Jul 18, 2015

The quotient rule says that for $f \left(x\right) = \frac{u}{v}$, the derivative is $f ' \left(x\right) = \frac{u ' v - u v '}{v} ^ 2$.

#### Explanation:

In this function $u = 1 + \sin \left(2 x\right)$

We can find $u '$ using the chain rule, if we have learned it yet, or by rewriting $1 + \sin \left(2 x\right) = 1 + 2 \sin x \cos x$ and using the product rule.

By chain rule:

$u ' = 0 + \cos \left(2 x\right) \cdot \frac{d}{\mathrm{dx}} \left(2 x\right)$

$= \cos \left(2 x\right) \cdot 2$

$= 2 \cos \left(2 x\right)$

By rewrite and product
For the derivative of a product there are a couple of ways of writing it. The difference is in the order in which we write things. I shall use:
$\left(f g\right) ' = f ' g + f g '$

The constant factor $2$, just hangs out in front.

$u = 1 + 2 \left[\sin x \cos x\right]$

$u ' = 0 + 2 \left[\left(\cos x\right) \cos x + \sin x \left(- \sin x\right)\right]$

$= 2 \left({\cos}^{2} x - {\sin}^{2} x\right)$ which you may remember from trigonometry is
$= 2 \cos \left(2 x\right)$

Similarly, with $v = 1 - \sin \left(2 x\right)$ we can get $v ' = - 2 \cos \left(2 x\right)$

So we have:

$f \left(x\right) = \frac{1 + \sin \left(2 x\right)}{1 - \sin \left(2 x\right)}$

and (derivatives are in brackets [ ])

$f ' \left(x\right) = \frac{\left[2 \cos \left(2 x\right)\right] \left(1 - \sin \left(2 x\right)\right) - \left(1 + \sin \left(2 x\right)\right) \left[- 2 \cos \left(2 x\right)\right]}{1 - \sin \left(2 x\right)} ^ 2$

We have found the derivative, but it is terribly messy. Let's do some algebra and maybe some trigonometry to make it less messy.

$f ' \left(x\right) = \frac{\left[2 \cos \left(2 x\right)\right] \left(1 - \sin \left(2 x\right)\right) - \left(1 + \sin \left(2 x\right)\right) \left[- 2 \cos \left(2 x\right)\right]}{1 - \sin \left(2 x\right)} ^ 2$

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

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

$= \frac{4 \cos \left(2 x\right)}{1 - \sin \left(2 x\right)} ^ 2$

Now, there are trigonometric identities we could apply, but none of them seem likely to simplify, so let's just stop here.

If you're curious here is a bit more:

Identities:

$1 - {\sin}^{2} \theta = {\cos}^{2} \theta$ but we can't use this in an obvious way

$\sin 2 x = 2 \sin x \cos x$ and

$\cos 2 x = {\cos}^{2} x - {\sin}^{2} x$

$= 1 - 2 {\sin}^{2} x$ initially looks like it might help, but if we're changing to $x$'s we get ${\left(1 - 2 \sin x \cos x\right)}^{2}$ in the denominator. That doesn't look promising either.

OK, that's enough. If we've overlooked something, we'll live with it.