Question

What are the first and second derivatives of `f(x)=ln((x-1)^2/(x+3))^(1/3)`?

Answer 1

`1/3[ln(x-1)^2 -ln(x+3)]=1/3[2ln(x-1)-ln(x+3)]=2/3 ln(x-1)-1/3ln(x+3)`
`[f'(x)=2/(3(x-1) ) -1/(3(x+3))]->[f''=-2/(3(x-1)^2)+1/(3(x+3)^2)]`

Explanation:

First use the properties of logarithms to simplify. Bring the exponent to the front and recall that the log of a quotient is the difference of the logs so once I dissolve it into simple logarithmic form then I find the derivatives. Once I have the first derivative then I bring up the `(x-1)` and`(x+3)` to the top and apply power rule to find the second derivative. Note that you can use chain rule as well but simplifying might be a bit harder and longer.