What is the derivative of ln(z^2/(z+1))?

1 Answer
Dec 31, 2015

We'll need chain rule, then quotient rule.

Explanation:

  • Chain rule: (dy)/(dx)=(dy)/(du)(du)/(dx)
  • Quotient rule: be y=f(x)/g(x), then y'=(f'g-fg')/g^2

Renaming f(x)=ln(u) and u=z^2/(z+1), we can differentiate it:

(dy)/(dz)=1/u(2z(z+1)-z^2(1))/(z+1)^2

Substituting u

(dy)/(dz)=((cancel(z+1))/z^cancel(2))*(2cancel(z)(z+1)-z^cancel(2)(1))/(z+1)^cancel(2)

(dy)/(dz)=(z+2)/(z(z+1))=color(red)((z+2)/(z^2+z))