AP Calculus AB and BC

Derivative of csc 2x: Chain Rule and Sign

The derivative of csc 2x is negative 2 times csc 2x times cot 2x. Cosecant differentiates to negative cosecant times cotangent, and the chain rule contributes the factor of 2 from the inner function.

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

The reciprocal derivative with an inner function

ddxcsc⁡(2x)=−csc⁡(2x)cot⁡(2x)⋅2\frac{d}{dx}\csc(2x) = -\csc(2x)\cot(2x)\cdot 2

Both the cosecant and the cotangent keep the SAME argument 2x2x. Only the multiplier changes.

Mirror of the secant version

ddxsec⁡2x=2sec⁡2xtan⁡2x,ddxcsc⁡2x=−2csc⁡2xcot⁡2x\frac{d}{dx}\sec 2x = 2\sec 2x\tan 2x, \qquad \frac{d}{dx}\csc 2x = -2\csc 2x\cot 2x

Identical structure, opposite sign, co-function partner. That pattern covers all four reciprocal derivatives at once.

Common mistakes

  • Changing the argument on one factor, as in −2csc⁡(2x)cot⁡(x)-2\csc(2x)\cot(x).
  • Losing the minus sign, which gives the derivative of −csc⁡2x-\csc 2x.
  • Pairing cosecant with tangent. It pairs with cotangent.

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 2x?

It is −2csc⁡2xcot⁡2x-2\csc 2x\cot 2x.

How does it compare with sec 2x?

ddxsec⁡2x=2sec⁡2xtan⁡2x\frac{d}{dx}\sec 2x = 2\sec 2x\tan 2x: same shape, opposite sign, co-function partner.