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

##### 1 Answer

The quotient rule says that for

#### Explanation:

In this function

We can find

By **chain rule:**

# = cos(2x)*2#

# = 2cos(2x)#

**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:

The constant factor

# = 2(cos^2x-sin^2x)# which you may remember from trigonometry is

# = 2cos(2x)#

Similarly, with

So we have:

and (derivatives are in brackets

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

# = (2cos(2x) (1-sin(2x))+2cos(2x)(1+sin(2x)))/(1-sin(2x))^2#

# = (2cos(2x) -2cos(2x)sin(2x)+2cos(2x)+2cos(2x)sin(2x))/(1-sin(2x))^2#

# = (4cos(2x))/(1-sin(2x))^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:**

# = 1-2sin^2x# initially looks like it might help, but if we're changing to#x# 's we get#(1-2sinxcosx)^2# in the denominator. That doesn't look promising either.

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