Question

What are the first and second derivatives of `g(x) = lncot^2x`?

Answer 1

see below

Explanation:

`g(x)=lncot^2x`

`g(x)=ln((cotx)^2)`

Use chain rule `(f(g(x)))'=f'(g(x)) * g'(x)`

`g'(x)=1/cot^2x * 2cot x * -csc^2x`

`g'(x)=(-2csc^2x)/cotx = -2csc^2 x tan x`

Use product rule`(fg)'=fg'+gf'` to find `g''(x)`

`f = -2csc^2x, g=tanx`

`f'=-4cscx * -cscx cot x, g'(x)=sec^2x`

`g''(x)=-2csc^2x sec^2x+4csc^2x cot x tanx`

`g''(x)=-2csc^2x sec^2x+4csc^2x`