What are the first and second derivatives of f(x)=ln((x-1)^2/(x+3))^(1/3) f(x)=ln((x1)2x+3)13?

1 Answer
Mar 7, 2016

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)13[ln(x1)2ln(x+3)]=13[2ln(x1)ln(x+3)]=23ln(x1)13ln(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.