Question

How do you find the derivative of `ln(x^3+3x)`?

Answer 1

`d/dxln(x^3+3x)=(3x^2+3)/(x^3+3x)`

Explanation:

We us the chain rule: `dy/dx=dy/(du)(du)/dx`

Let `y=ln(x^3+3x)`, and
Let `u=x^3+3x`

`y=ln(x^3+3x)`
`:. y=lnu`
`:. dy/(du)=1/u`

`u=x^3+3x`
`:. (du)/dx=3x^2+3`

So using the chain rule we have:
`dy/dx=dy/(du)(du)/dx`
`:. dy/dx=1/u(3x^2+3)`
`:. dy/dx=(3x^2+3)/(x^3+3x)`