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Inverse trig derivatives

Derivative of arccot x: Formula, Proof and Examples

The derivative of arccot x is −1/(1 + x²). See the proof from cot y = x, the link to arctan, chain rule examples and a graph.

Inverse trig derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\operatorname{arccot} x = -\frac{1}{1 + x^2}
Derivative
-\dfrac{1}{1 + x^2} for every real x
Chain rule form
\dfrac{d}{dx}\operatorname{arccot} u = -\dfrac{u'}{1 + u^2}
Domain of arccot x
all real numbers
Range of arccot x
(0, \pi)
Key identity
\operatorname{arccot} x = \frac{\pi}{2} - \arctan x
On this page
  1. Graph of arccot x and its derivative
  2. Proof from cot y = x
  3. Proof from arctan
  4. A note on conventions
  5. Chain rule version and examples
  6. Common mistakes

The derivative of \operatorname{arccot} x is -\dfrac{1}{1 + x^2}. It is the derivative of arctan x with a minus sign, and it holds for every real number.

\frac{d}{dx}\,\operatorname{arccot} x = -\frac{1}{1 + x^2}

Graph of arccot x and its derivative

Graph of y = arccot x and its derivative
The inverse cotangent falls from \pi toward 0. Its slope is -1 at x = 0 and fades to 0 far from the origin.

Proof from cot y = x

Let y = \operatorname{arccot} x, so \cot y = x with 0 < y < \pi. Differentiate with implicit differentiation, using \frac{d}{dy}\cot y = -\csc^2 y (see the derivative of cot x):

-\csc^2 y \cdot \frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{1}{\csc^2 y}

The identity \csc^2 y = 1 + \cot^2 y = 1 + x^2 finishes it:

\frac{d}{dx}\operatorname{arccot} x = -\frac{1}{1 + x^2}

Proof from arctan

With the range (0, \pi), arccot and arctan are related by

\operatorname{arccot} x = \frac{\pi}{2} - \arctan x

The derivative of the constant \frac{\pi}{2} is 0, so the derivative of arccot is just minus the derivative of arctan. This identity is also how the graph above is drawn.

A note on conventions

Some books and software define \operatorname{arccot} x = \arctan\frac{1}{x}, which has range \left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right] and jumps at x = 0. The derivative still comes out the same wherever the function is defined:

\frac{d}{dx}\arctan\frac{1}{x} = \frac{-1/x^2}{1 + 1/x^2} = -\frac{1}{x^2 + 1}

So the formula is safe to use either way. Only the values of the function differ.

Chain rule version and examples

\frac{d}{dx}\operatorname{arccot} u = -\frac{u'}{1 + u^2}
Function Derivative
\operatorname{arccot}(3x) -\frac{3}{1 + 9x^2}
\operatorname{arccot}(x^2) -\frac{2x}{1 + x^4}
\operatorname{arccot}(\ln x) -\frac{1}{x(1 + \ln^2 x)}
\operatorname{arccot}(e^x) -\frac{e^x}{1 + e^{2x}}

Worked example. Differentiate f(x) = x\operatorname{arccot} x with the product rule:

f'(x) = \operatorname{arccot} x - \frac{x}{1 + x^2}

At x = 1: \operatorname{arccot} 1 = \frac{\pi}{4}, so f'(1) = \frac{\pi}{4} - \frac{1}{2} \approx 0.285.

Common mistakes

  1. Writing -\csc^2 x. That is the derivative of \cot x, not of the inverse.
  2. Losing the minus sign. Arccot decreases, so its slope must be negative.
  3. Forgetting u'. \operatorname{arccot}(5x) gives -\frac{5}{1 + 25x^2}.
  4. Assuming the integral needs arccot. Since the derivatives differ only by sign, \int \frac{dx}{1 + x^2} is usually written as \arctan x + C. The answer -\operatorname{arccot} x + C is also correct, because the two differ by a constant.

All six inverse trig derivatives come in pairs: arcsin and arccos, arctan and arccot, arcsec and arccsc. The second of each pair has a minus sign. Practise them with inverse trig derivative problems.

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Frequently asked questions

What is the derivative of arccot x?
The derivative of \operatorname{arccot} x (also written \cot^{-1} x) is -\frac{1}{1 + x^2}.

What is the derivative of arccot(2x)?
By the chain rule it is -\frac{2}{1 + 4x^2}.

Is arccot x equal to arctan(1/x)?
Only for x > 0. For x < 0 they differ by \pi. Either way, where both are defined they have the same derivative, -\frac{1}{1 + x^2}.

What is the derivative of arctan x + arccot x?
It is 0, since the sum is the constant \frac{\pi}{2}.

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Related derivatives

Derivative of arctan x \frac{d}{dx}\,\arctan x = \frac{1}{1 + x^2} The derivative of arctan x is 1/(1 + x²). See the proof from tan y = x, chain rule examples like arctan(2x) and arctan(x²), and a graph. Derivative of arccos x \frac{d}{dx}\,\arccos x = -\frac{1}{\sqrt{1 - x^2}} The derivative of arccos x is −1/√(1 − x²). See the proof, why it is the negative of the arcsin derivative, chain rule examples and a graph. Derivative of arcsec x \frac{d}{dx}\,\operatorname{arcsec} x = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}} The derivative of arcsec x is 1/(|x|√(x² − 1)). See where the absolute value comes from, a proof via arccos(1/x), and chain rule examples.