Deriving The Arccotangent Derivative

Why the Arccotangent Derivative Is Negative

The inverse trigonometric derivatives share a common method: represent the inverse function by an equation involving the original trigonometric function, then differentiate that equation. For arccotangent, the defining equation is \(\cot(y)=x\). The sign in the resulting derivative is important: as \(x\) increases, the principal arccotangent value decreases.

The sign is not a convention that can be chosen independently of the inverse function. It follows from the fact that cotangent decreases on its principal interval. We will make that interval explicit, derive the formula, and check it at several exact inputs.

The Principal Range Makes Cotangent Invertible

Cotangent is not one-to-one on all of its domain because it repeats its values periodically. To define an inverse function, we restrict cotangent to the interval \((0,\pi)\). On that interval, \(\sin y\ne 0\), so \(\cot y=\cos y/\sin y\) is defined.

Using the trigonometric derivative, \(\frac{d}{dy}\cot y=-\csc^2 y\). Since \(\csc^2 y>0\) wherever it is defined, this derivative is strictly negative on \((0,\pi)\). Therefore cotangent is strictly decreasing there. As \(y\) approaches \(0\) from the right, \(\cot y\) tends to \(+\infty\); as \(y\) approaches \(\pi\) from the left, \(\cot y\) tends to \(-\infty\). Continuity and strict decrease show that cotangent takes each real value exactly once on this interval. This establishes the stated inverse and its domain.

The interval is open: neither endpoint is included, and cotangent is not defined at either endpoint. Nevertheless, every real \(x\) has an arccotangent value strictly between \(0\) and \(\pi\).

Deriving the Formula

Let \(y=\arccot x\). By the definition and principal-value convention, \(y\in(0,\pi)\) and \(\cot y=x\). Cotangent is differentiable on this interval, and its derivative is never zero there. The inverse-function derivative rule therefore applies: the inverse is differentiable at every real \(x\). We can obtain its derivative directly by differentiating the defining equation with respect to \(x\).

Because \(y\) depends on \(x\), differentiating the left side requires the chain rule. The derivative of \(\cot y\) with respect to \(x\) is \(-\csc^2 y\,\frac{dy}{dx}\), while the derivative of \(x\) is \(1\). Hence

Since \(\csc^2 y\) is positive and finite for \(y\in(0,\pi)\), we can divide by it and isolate the derivative:

The identity \(\csc^2 y=1+\cot^2 y\), together with \(\cot y=x\), replaces the trigonometric expression involving \(y\) by one involving \(x\). Since \(y=\arccot x\), this gives the derivative formula.

Why the Sign and Domain Follow from the Derivation

The negative sign has a geometric meaning. Cotangent decreases as its angle increases on \((0,\pi)\). Its inverse must therefore decrease as its input increases: larger values of \(x\) correspond to smaller principal angles \(y\). The formula confirms this, because \(1+x^2>0\) for every real \(x\), so \(-1/(1+x^2)<0\).

The derivative’s domain is all of \(\mathbb{R}\), matching the domain of arccotangent under this convention. There is no real \(x\) for which the denominator vanishes: \(1+x^2\geq 1\). The output range remains \((0,\pi)\). These facts concern different things: the domain tells us which inputs are allowed, while the range tells us which principal angles are returned.

Worked Examples

A Common Sign Error

A frequent error is to differentiate \(\cot y\) as though its derivative were \(+\csc^2 y\). That incorrect sign changes the final answer. For example, the mistaken calculation would be

The first implication is wrong: the correct derivative is \(\frac{d}{dy}\cot y=-\csc^2 y\). The corrected calculation is

The positive answer also conflicts with the behavior of the inverse: it would say that arccotangent increases, even though cotangent is decreasing on the chosen principal interval. Checking the sign against that behavior is a useful way to catch the error.

What the Formula Tells Us

The derivative formula packages both the local rate of change and the domain information. Its denominator is always positive, so the slope is always negative. Also, as \(|x|\) grows, \(1+x^2\) grows, and the magnitude of the slope becomes smaller. Thus the inverse function changes less rapidly at inputs far from zero than it does near zero.

These observations do not replace the derivation: they are consequences of it. The crucial ingredients were the specified principal interval, the inverse relation \(\cot y=x\), the chain rule, and the identity \(\csc^2 y=1+\cot^2 y\). If a different convention for arccotangent were used, its range would need to be stated and the inverse relation checked under that convention.