Question

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

Answer 1

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))`