Differentiate ln(ln(2-y)))?

Question

Hard time with 2 ln's

Answer 1

It's just the chain rule.

[ln(ln(2 - y))]'

1/ln(2 - y) × [ln(2 - y)]'

1/ln(2 - y) × 1/(2 - y) × (2 - y)'

1/ln(2 - y) × 1/(2 - y) × (-1)

Put it all together and you get:

-1/[(2 - y)(ln(2 - y))]