Derivative of cot(x)
The derivative of the cotangent is $\dfrac{d}{dx}\cot x = -\csc^2 x = -\dfrac{1}{\sin^2 x} = -(1+\cot^2 x)$, for $x \neq k\pi$.
$$ \frac{d}{dx}\left(\cot x\right) = -\csc^2 x $$
See the derivative of cot(x) step by step in the calculator →
📈 Graph of cot(x) and its derivative
🔍 Why is the derivative of cot(x) equal to $-\csc^2 x$?
With $\cot x = \frac{\cos x}{\sin x}$ and the quotient rule:
$$ (\cot x)' = \frac{-\sin x\cdot\sin x - \cos x\cdot\cos x}{\sin^2 x} = -\frac{\sin^2 x + \cos^2 x}{\sin^2 x} = -\frac{1}{\sin^2 x}. $$
Compare with $(\tan x)' = \frac{1}{\cos^2 x}$: swapping sine and cosine and adding a minus sign – the usual pattern for the "co"-functions.
🔗 Derivative of cot(x) with an inner function
If the argument is a function $u(x)$, the chain rule adds the inner derivative:
$$ \frac{d}{dx}\cot\big(u(x)\big) = -u'(x)\csc^2\big(u(x)\big) $$
✏️ Worked examples
⚠️ Common mistakes
Forgetting the minus sign.
Writing $-\csc x$ instead of $-\csc^2 x$.
❓ Frequently asked questions
What is the derivative of 1/tan(x)?
$\frac{1}{\tan x} = \cot x$, so the derivative is $-\csc^2 x = -\frac{1}{\sin^2 x}$.
What is the derivative of cot(2x)?
$-2\csc^2(2x)$.
Is cot(x) ever increasing?
No. Its derivative is negative everywhere on its domain.