Question

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

Answer 1

There are a couple of approaches, all of which will involve the chain rule, which I assume you are familiar with.

Method 1 - Use the chain rule (directly)

`y = ln(ln(3x^3))`
`:. dy/dx = 1/(ln(3x^3)) * 1/(3x^3) * 9x^2`
`:. dy/dx = 3/(xln(3x^3))`

Method 2 - Use the chain rule (with substitution)

`{ ("Let "u =3x^3, => (du)/dx=9x^2), ("Let "v= lnu, => (dv)/(du)=1/u), ("Then "y=lnv =ln(lnu)=ln(ln(3x^3))", => dy/(dv)=1/v) :}`

By the Chain Rule:

`dy/dx = dy/(dv) * (dv)/(du) * (du)/dx`
`:. dy/dx = 1/v * 1/u * 9x^2`
`:. dy/dx = 1/lnu * 1/(3x^3) * 9x^2`
`:. dy/dx = 1/ln(3x^3) * 3/x`
`:. dy/dx = 3/(xln(3x^3))`

Method 3 - Take Exponents and differentiate Implicitly

`y = ln(ln(3x^3))`
`:. e^y = ln(3x^3)`
`:. e^ydy/dx = 1/(3x^3)*9x^2`
`:. ln(3x^3)dy/dx = 3/x`
`:. dy/dx = 3/(xln(3x^3))`