Question

How do you differentiate `y=ln(9x+4)^2`?

Answer 1

`dy/dx=(18ln(9x+4))/(9x+4)`

Explanation:

The chain rule states that `dy/dx=dy/(du)(du)/dx` for some function `u` of `x`.

In this problem, let `u=ln(9x+4)`, so `y=u^2`. Therefore, `dy/dx=d/(du)[u^2]times(du)/dx`
`dy/dx=2utimes(du)/dx`

And substitute back in for `u`.
`dy/dx=2ln(9x+4)times(du)/dx`

So now we need to find `(du)/dx`, we can use the chain rule again. Now say `(du)/dx=(du)/(dv)(dv)/dx`

Let `v=9x+4` so `u=lnv`
`(du)/dx=d/(dv)[lnv]times(dv)/dx`
`(du)/dx=1/vtimes(dv)/dx`

And substitute back in for `v`.
`(du)/dx=1/(9x+4)times(dv)/dx`

And finally we need to find `(dv)/dx`.
`(dv)/dx=d/dx[9x+4]`
`(dv)/dx=9`

And finally plug everything back in.
`(du)/dx=1/(9x+4)times9=9/(9x+4)`
`dy/dx=2ln(9x+4)times9/(9x+4)=(18ln(9x+4))/(9x+4)`