How do you differentiate #f(x)=ln(sqrt(sin(x^2-3x)))# using the chain rule?

1 Answer
Mar 4, 2016

Answer:

#f'(x)=1/(sqrt(sin(x^2-3x))) *1/(2sqrt(sin(x^2-3x))) cos (x^2-3x)(2x-3)#
#f'(x)=((2x-3)cos(x^2-3x))/(2sin(x^2-3x)) = 1/2 (2x-3)cot(x^2-3x)#

Explanation:

You have four functions in here. Start with the #ln# and keep everything else the same then move to the #sqrt#, then sine and then #x^2-3x#. Then simplify note that# cos(x^2-3x)/(sin(x^2-3x))=cot(x^2-3x)#