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

1 Answer
Dec 25, 2015

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)