Question

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

Answer 1

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))`