Question

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

Answer 1

The quotient rule says that for `f(x) = u/v`, the derivative is `f'(x) = (u'v-uv')/v^2`.

Explanation:

In this function `u = 1+sin(2x)`

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

By chain rule:

`u' = 0+cos(2x)*d/dx(2x)`

`= 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:
`(fg)' = f'g+fg'`

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

`u = 1+2[sinxcosx]`

`u' = 0+ 2[(cosx)cosx+sinx(-sinx)]`

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

Similarly, with `v = 1-sin(2x)` we can get `v'=-2cos(2x)`

So we have:

`f(x) = (1+sin(2x))/(1-sin(2x))`

and (derivatives are in brackets `[` `]`)

`f'(x) = ([2cos(2x)] (1-sin(2x))-(1+sin(2x))[-2cos(2x)])/(1-sin(2x))^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'(x) = ([2cos(2x)] (1-sin(2x))-(1+sin(2x))[-2cos(2x)])/(1-sin(2x))^2`

`= (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-sin^2 theta = cos^2 theta` but we can't use this in an obvious way

`sin2x = 2sinxcosx` and

`cos2x = cos^2x-sin^2x`

`= 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.