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

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Answer 1 · 2015-12-14T03:30:42

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

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

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.

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

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

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