AP Calculus AB and BC

Derivative of cot 2x: Chain Rule and Sign

Q: Where do the two parts come from?

The minus and the \csc^{2} come from differentiating cotangent; the 2 comes from the chain rule on the inner 2x.

The derivative of cot 2x is negative 2 times csc squared of 2x. Two separate things produce the answer: cotangent differentiates to negative csc squared, and the chain rule multiplies by 2 for the inner function.

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

Two factors, two sources

\frac{d}{dx}\cot(2x) = -\csc^{2}(2x)\cdot 2 = -2\csc^{2}2x

The minus sign comes from cotangent being a co-function; the $2$ comes from the chain rule. Losing one and keeping the other is the usual error.

Domain and asymptotes

$\cot 2x$ is undefined wherever $\sin 2x = 0$ , that is at every multiple of $\frac{\pi}{2}$ . Doubling the input halves the spacing of the asymptotes compared with $\cot x$ .

Common mistakes

Answering $2\csc^{2}2x$ with no minus sign.
Answering $-\csc^{2}2x$ , dropping the chain rule factor.
Writing $-2\csc^{2}x$ , differentiating the inner function into the argument.

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 2x?

It is $-2\csc^{2}2x$ .

Where do the two parts come from?

The minus and the $\csc^{2}$ come from differentiating cotangent; the $2$ comes from the chain rule on the inner $2x$ .