Question

How do you differentiate `f(x)=(ln(-x+3))^2` using the chain rule.?

Answer 1

The answer is `=(-2ln(-x+3))/(-x+3)`

Explanation:

We apply the chain rule.
`(x^n)'=nx^(n-1)`
`(lnx)'=1/x`
`(ax+b)'=a`

`((ln(-x+3))^2)'=2ln(-x+3) * (ln(-x+3))' * (-x+3)'`

`= 2ln(-x+3)*(1/(-x+3))(-1)`

`=(-2ln(-x+3))/(-x+3)`