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))#