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

1 Answer
Jul 18, 2015

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.