AP Calculus AB and BC

Derivative of csc^2 x: Chain Rule

The derivative of csc squared x is negative 2 times csc squared x times cot x. The power rule handles the outer square, giving twice csc x, and the chain rule multiplies by the derivative of cosecant, which is negative csc x cot x.

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

Two layers

ddx(csc⁡x)2=2csc⁡x⋅(−csc⁡xcot⁡x)=−2csc⁡2xcot⁡x\frac{d}{dx}\left(\csc x\right)^{2} = 2\csc x\cdot\left(-\csc x\cot x\right) = -2\csc^{2}x\cot x

The two cosecants combine into a square, which is why the answer looks like the original function times a cotangent.

Read backwards it is an antiderivative

Since ∫csc⁡2x dx=−cot⁡x+C\int \csc^{2}x\,dx = -\cot x + C, this page and that antiderivative are the same fact in opposite directions. Keeping them apart is what the exam actually tests.

Common mistakes

  • Dropping a minus sign. There is exactly one, from the cosecant derivative.
  • Pairing cosecant with tangent. It pairs with cotangent.
  • Reading csc⁡2x\csc^{2}x as csc⁡(x2)\csc\left(x^{2}\right).

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^2 x?

It is −2csc⁡2xcot⁡x-2\csc^{2}x\cot x.

Where do the two factors come from?

The 22 from the outer power rule, and −csc⁡xcot⁡x-\csc x\cot x from differentiating cosecant.