Find F"(x) if f(x) = cot (x)

Question

asked Aug 19, 2020 38.4k views

1 Answer

Mnacos · Answer 1 · 2020-08-26T02:38:52+0000

$f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}$

If you don't know the first derivative of
$\cot$ , but you do for
$\sin$ and
$\cos$ , you can derive the former via the quotient rule:

$\cot x=(\cos x)/(\sin x)$

$\implies(\cot x)'=(\sin x(-\sin x)-\cos x(\cos x))/(\sin^2x)=-\frac1{\sin^2x}=-\csc^2x$

or if you know the derivative of
$\tan$ :

$\cot x=\frac1{\tan x}$

$\implies(\cot x)'=-(\tan x)^(-2)\sec^2x=-(\sec^2x)/(\tan^2x)=-\frac{\frac1{\cos^2x}}{(\sin^2x)/(\cos^2x)}=-\frac1{\sin^2x}=-\csc^2x$

As for the second derivative, you can use the power/chain rules:

$(-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x$

or if you don't know the derivative of
$\csc$ ,

$\csc x=\frac1{\sin x}$

$\implies(-\csc^2x)'=\left(-(\sin x)^(-2)\right)'=2(\sin x)^(-3)(\sin x)'=(2\cos x)/(\sin^3x)$

which is the same as the previous result since

$\csc^2x\cot x=\frac1{\sin^2x}(\cos x)/(\sin x)=(\cos x)/(\sin^3x)$