How do I find the derivative of #f(x) = ln sqrt ((2x-8 )/ (3x+3))#?

1 Answer
Jan 12, 2016

We'll need chain rule and quotient rule.

Explanation:

  • Chain rule: #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

  • Quotient rule: #(a/b)'=(a'b-ab')/b^2#

Renaming #u=sqrt(v)# and #v=(2x-8)/(3x+3)#, we can proceed.

#(dy)/(dx)=1/u1/(2v^(1/2))((2(3x+3)-3(2x-8))/(3x+3)^2)#

Substituting #u#, #v# and simplifying:

#(dy)/(dx)=1/(2(sqrt(v))^2)(cancel(6x)+6-cancel(6x)+24)/(3x+3)^2#

#(dy)/(dx)=15/((3x+3)^(3/2)(2x-8)^(1/2))#