#f(x) = sin(log(x)) and y = f((2x +3) / (3 - 2x))#
I will assume that the base of #log(x)# is #e# and I will use #ln(x)# in my answer.
#y# is a composition of functions, so let's add some notation to make this more obvious.
Let:
#p(x) = sin(x)#
#q(x)= ln(x)#
#u(x) = 2x + 3#
#v(x) = 3- 2x#
then:
#y(x) = p(q((u(x))/(v(x))))#
We could attack this directly, but a property of logs can simplify it a bit:
#q((u(x))/(v(x))) = ln((u(x))/(v(x))) = ln(u(x)) - ln(v(x))#
#u'(x) = 2#
#v'(x) = -2#
#q'(x) = (u'(x))/(u(x)) - (v'(x))/(v(x)) = 2/(2x + 3) - (-2)/(3- 2x)#
#= 2/(2x + 3) - 2/(2x- 3) = (4x - 6 -(4x + 6))/(4x^2 - 9)#
#= -(12)/(4x^2 - 9)#
Applying the Chain Rule to y(x):
#y'(x) = p'(q(x)) * q'(x)#
#p'(x) = cos(x)#
so:
#y'(x) = cos(ln((2x + 3)/(3- 2x))) * (-(12)/(4x^2 - 9))#
# = -(12*cos(ln((2x + 3)/(3- 2x))))/(4x^2 - 9)#
SEE: https://www.khanacademy.org/math/ap-calculus-ab/ab-derivatives-advanced/ab-diff-log/v/chain-rule-with-triple-composition
