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.