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))