[tex]f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}[/tex]If you don't know the first derivative of [tex]\cot[/tex], but you do for [tex]\sin[/tex] and [tex]\cos[/tex], you can derive the former via the quotient rule:[tex]\cot x=\dfrac{\cos x}{\sin x}[/tex][tex]\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x[/tex]or if you know the derivative of [tex]\tan[/tex]:[tex]\cot x=\dfrac1{\tan x}[/tex][tex]\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[/tex]As for the second derivative, you can use the power/chain rules:[tex](-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x[/tex]or if you don't know the derivative of [tex]\csc[/tex],[tex]\csc x=\dfrac1{\sin x}[/tex][tex]\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}[/tex]which is the same as the previous result since[tex]\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}[/tex]