AP Calculus AB and BC

Derivative of csc x: Answer, Proof, and Mistakes

The derivative of csc x with respect to x is negative csc x times cot x. Writing csc x as sin x to the negative one and applying the chain rule produces it directly. Like every co-function derivative it carries a minus sign, and it is the exact mirror of the derivative of sec x.

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

How to differentiate csc x

Write csc⁡x=(sin⁡x)−1\csc x = \left(\sin x\right)^{-1} and use the chain rule with the power rule.

ddx(sin⁡x)−1=−(sin⁡x)−2⋅cos⁡x=−cos⁡xsin⁡2x\frac{d}{dx}\left(\sin x\right)^{-1} = -\left(\sin x\right)^{-2}\cdot\cos x = \frac{-\cos x}{\sin^{2} x}

Split that fraction into 1sin⁡x\frac{1}{\sin x} times cos⁡xsin⁡x\frac{\cos x}{\sin x} to recognise the standard form.

ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\csc x = -\csc x\cot x

Why it mirrors the derivative of sec x

The two reciprocal derivatives have identical structure, differing only by the minus sign and by which pair of functions appears.

ddxsec⁡x=sec⁡xtan⁡x,ddxcsc⁡x=−csc⁡xcot⁡x\frac{d}{dx}\sec x = \sec x\tan x, \qquad \frac{d}{dx}\csc x = -\csc x\cot x

Each derivative reproduces its own function times the matching tangent-family function. Remembering that shape is more reliable than memorising four separate expressions.

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

Topic 2.7 covers the trigonometric derivatives on both AB and BC. Reversed, it gives ∫csc⁡xcot⁡x dx=−csc⁡x+C\int \csc x\cot x\,dx = -\csc x + C, a basic antiderivative.

With an inner function you also need the chain rule: ddxcsc⁡(3x)=−3csc⁡(3x)cot⁡(3x)\frac{d}{dx}\csc(3x) = -3\csc(3x)\cot(3x).

Common mistakes with the derivative of csc x

  • Losing the minus sign, which turns the answer into the derivative of −csc⁡x-\csc x.
  • Writing −csc⁡xtan⁡x-\csc x\tan x. Cosecant pairs with cotangent, not tangent.
  • Answering −sin⁡−2x-\sin^{-2}x and stopping. That is correct but incomplete for matching an answer key, since the cos⁡x\cos x from the chain rule is missing.

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 csc x?

It is −csc⁡xcot⁡x-\csc x\cot x.

How is it related to the derivative of sec x?

They mirror each other: sec⁡xtan⁡x\sec x\tan x against −csc⁡xcot⁡x-\csc x\cot x. Same structure, opposite sign, co-function partners.

What is the antiderivative of csc x cot x?

It is −csc⁡x+C-\csc x + C, the derivative read backwards.

What is the derivative of csc(3x)?

It is −3csc⁡(3x)cot⁡(3x)-3\csc(3x)\cot(3x); the chain rule contributes the factor of 33.