Question

What are the first and second derivatives of `g(x) = lntanx`?

Answer 1

`g'(x)=sec^2x/tanx`

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

Explanation:

Know that `d/dx[ln(u)]=(u')/u`.

Therefore,

`g'(x)=(d/dx[tanx])/tanx=sec^2x/tanx`

`g''(x)=(tanxd/dx[sec^2x]-sec^2xd/dx[tanx])/tan^2x`

Find each derivative separately.

Chain rule:
`d/dx[sec^2x]=2secxd/dx[secx]=2secx(secxtanx)=2sec^2xtanx`

`d/dx[tanx]=sec^2x`

Plug these back in.

`g''(x)=(2sec^2xtan^2x-sec^4x)/tan^2x`

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