Question

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

Answer 1

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

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:

$\cot x=\dfrac{\cos x}{\sin x}$

$\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x$

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

$\cot x=\dfrac1{\tan x}$

$\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x$

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 $\csc$,

$\csc x=\dfrac1{\sin x}$

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

which is the same as the previous result since

$\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}$