How do you differentiate #f(x)=sqrtcos(lnx^2)# using the chain rule?

1 Answer
Nov 9, 2015

Answer:

found: #=1/x*(-sin(ln(x^2)))/sqrt(cos(ln(x^2))#

Explanation:

Here, basically, you have 4 functions one inside the other! Starting for the most "external" you have:
the square root or #()^(1/2)#, the #cos#, then the #ln# and finally the square #()^2#.
To derive you start from the outer one (the square root) and subsequently multiply by each derivative up the inner one (colored).
So:
given: #f(x)=(cos(ln(x^2)))^(1/2)# you get:

#f'(x)=1/2(cos(ln(x^2)))^(1/2-1)*color(red)(-sin(ln(x^2)))*color(blue)(1/x^2)*color(green)(2x)=#

rearranging:

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

#=1/x*(-sin(ln(x^2)))/sqrt(cos(ln(x^2))#