AP Calculus AB and BC

Derivative of cot x: Answer, Proof, and Mistakes

The derivative of cot x with respect to x is negative csc squared x. It follows from the quotient rule applied to cos x over sin x, where the Pythagorean identity collapses the numerator to negative one. The minus sign matches the pattern that every co-function derivative is negative.

ddx[cot⁡x]=−csc⁡2x\frac{d}{dx}\left[\cot x\right] = -\csc^{2} x

How to differentiate cot x

Write cot⁡x\cot x as cos⁡xsin⁡x\frac{\cos x}{\sin x} and apply the quotient rule.

ddxcos⁡xsin⁡x=−sin⁡x⋅sin⁡x−cos⁡x⋅cos⁡xsin⁡2x\frac{d}{dx}\frac{\cos x}{\sin x} = \frac{-\sin x\cdot\sin x - \cos x\cdot\cos x}{\sin^{2} x}

The numerator is −(sin⁡2x+cos⁡2x)-\left(\sin^{2} x + \cos^{2} x\right), which the Pythagorean identity turns into −1-1.

ddxcot⁡x=−1sin⁡2x=−csc⁡2x\frac{d}{dx}\cot x = \frac{-1}{\sin^{2} x} = -\csc^{2} x

The co-function sign pattern

Every derivative in the trig table whose function starts with "co" carries a minus sign. That is one fact instead of six, and it is the fastest way to catch a sign slip.

  • ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x = \sec^{2} x against ddxcot⁡x=−csc⁡2x\frac{d}{dx}\cot x = -\csc^{2} x
  • ddxsin⁡x=cos⁡x\frac{d}{dx}\sin x = \cos x against ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin x
  • ddxsec⁡x=sec⁡xtan⁡x\frac{d}{dx}\sec x = \sec x\tan x against ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\csc x = -\csc x\cot x

Where the derivative of cot x shows up on the AP exam

Trigonometric derivatives are Topic 2.7, on both AB and BC. Read backwards, this result is also the antiderivative fact ∫csc⁡2x dx=−cot⁡x+C\int \csc^{2} x\,dx = -\cot x + C, which is where it earns most of its Unit 6 marks.

Note the domain. Cotangent is undefined wherever sin⁡x=0\sin x = 0, so the derivative does not exist at integer multiples of π\pi either.

Common mistakes with the derivative of cot x

  • Dropping the minus sign and writing csc⁡2x\csc^{2} x. That is the derivative of −cot⁡x-\cot x.
  • Writing −sec⁡2x-\sec^{2} x. Secant belongs to tangent; cosecant belongs to cotangent.
  • Writing −csc⁡2x-\csc^{2} x as −csc⁡(x2)-\csc\left(x^{2}\right). The square is on the function value, not on the input.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

What is the derivative of cot x?

It is −csc⁡2x-\csc^{2} x, which is −1sin⁡2x\frac{-1}{\sin^{2} x}.

Why does the derivative of cot x have a minus sign?

The quotient rule produces −(sin⁡2x+cos⁡2x)-\left(\sin^{2} x + \cos^{2} x\right) on top, which is −1-1. Every co-function derivative inherits a minus for the same structural reason.

What is the antiderivative of csc^2 x?

It is −cot⁡x+C-\cot x + C, which is this derivative read in reverse.

Where is cot x not differentiable?

Wherever sin⁡x=0\sin x = 0, that is at every integer multiple of π\pi, because the function itself is undefined there.