Question

How do I find the derivative of `ln sqrt(x^2-4)`?

Answer 1

We'll need the chain rule, which states that `(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)`

Explanation:

Renaming `u=sqrt(v)` and `v=x^2-4`, let's do it step-by-step:

`(dy)/(du)=1/u`

`(du)/(dv)=1/(2sqrt(v))`

`(dv)/(dx)=2x`

Aggregating:

`(dy)/(dx)=(2x)/(2usqrt(v))`

Substituting `u` and then `v` afterwards:

`(dy)/(dx)=(2x)/(2(sqrt(x^2-4))(sqrt(x^2-4)))=(cancel(2)x)/(cancel(2)(x^2-4))`

`(dy)/(dx)=x/(x^2-4)`