Why Arctangent Has a Derivative on the Whole Real Line
The tangent function is not one-to-one on its entire domain, so it does not have a single inverse there. Restricting tangent to the interval \( (-\pi/2,\pi/2) \) makes it one-to-one. On that interval, its outputs cover every real number. Its inverse is the arctangent function, written \( \arctan x \).
To find the derivative of \( \arctan x \), we can write the inverse relationship as an equation involving tangent, then differentiate that equation implicitly. This approach also reveals why the resulting derivative formula is valid for every real input.
Why Arctangent Accepts Every Real Input
To define an inverse function for every real input \(x\), the restricted tangent function must produce every real number exactly once. On \( (-\pi/2,\pi/2) \), tangent is continuous and has derivative \( \sec^2 y \). Since \( \sec^2 y>0 \) throughout this interval, tangent is strictly increasing there.
At the left endpoint, as \(y\) approaches \(-\pi/2\) from within the interval, \( \sin y\) approaches \(-1\) and \( \cos y\) approaches \(0\) through positive values. Therefore \( \tan y=\sin y/\cos y\) decreases without bound. At the right endpoint, \( \sin y\) approaches \(1\) while \( \cos y\) approaches \(0\) through positive values, so tangent increases without bound. Continuity and the Intermediate Value Theorem then show that tangent takes every real value on the interval. Strict increase shows that it takes each value only once.
This establishes that \( \arctan x \) is meaningful for every real \(x\). To establish differentiability, we also need the slope of tangent at the corresponding angle not to be zero. In fact, that slope is \( \sec^2 y>0 \) throughout the interval, so the inverse function derivative formula applies at every point in the range.
Deriving the Derivative by Implicit Differentiation
Let \(y=\arctan x\). By the definition of arctangent, \( \tan y=x \), where \(y\) lies in \( (-\pi/2,\pi/2) \). Here \(y\) depends on \(x\). Differentiate both sides with respect to \(x\), using the chain rule on the left:
Because \(y=y(x)\), differentiating \( \tan y\) gives \( \sec^2 y\,\frac{dy}{dx}\), not just \( \sec^2 y\). Thus,
The factor \( \sec^2 y\) is positive, so we can divide by it:
The identity \( \sec^2 y=1+\tan^2 y\) lets us replace the remaining expression involving \(y\) with the input \(x\). Since \( \tan y=x\),
Why the Formula Is Defined for Every Real Input
There are two related domain questions. First, \( \arctan x\) exists for every real \(x\), because the restricted tangent function has range \( \mathbb{R} \). Second, the derivative formula is defined at every real \(x\), because its denominator cannot be zero:
Consequently, the formula has no real inputs at which it is undefined. This matches the inverse-function reasoning: at every angle in the restricted interval, \( \sec^2 y>0\), so the tangent function has a nonzero derivative there.
The domain of the inverse function and the domain of its derivative formula should not be confused with the interval of its outputs. Every real number can be an input to arctangent, but the resulting angle always lies strictly between \(-\pi/2\) and \(\pi/2\).
Worked Examples
Worked Example: Find the Derivative at Input Zero
Let \(y=\arctan x\). The derivative formula gives \( y'=\frac{1}{1+x^2}\). At \(x=0\), substitute the input:
To check this value using the defining relation, \( \tan y=x\). Differentiating gives \( \sec^2 y\,y'=1\). At \(x=0\), the relation gives \( \tan y=0\), so the identity \( \sec^2 y=1+\tan^2 y\) gives \( \sec^2 y=1+0^2=1\). Therefore \(y'=1/1=1\), agreeing with the formula.
Worked Example: Find the Derivative at Input Two
For \(y=\arctan x\), evaluate the derivative at \(x=2\):
Check this result from the implicit equation. When \(x=2\), \( \tan y=2\), so \( \sec^2 y=1+\tan^2 y=1+2^2=5\). Differentiating \( \tan y=x\) gives \( \sec^2 y\,y'=1\), hence \(5y'=1\) and \(y'=1/5\). This matches the calculated value.
Worked Example: Find the Derivative at Input Negative Three
Evaluate the derivative of \( \arctan x\) at \(x=-3\):
For the check, the defining equation at this input is \( \tan y=-3\). The trigonometric identity gives \( \sec^2 y=1+\tan^2 y=1+(-3)^2=10\). Implicit differentiation gives \(10y'=1\), so \(y'=1/10\), as required. The negative input produces a positive derivative because the denominator contains \(x^2\).
A Chain-Rule Pitfall
A common mistake is to differentiate \( \tan y=x\) while treating \(y\) as though it were independent of \(x\). That loses the chain-rule factor \(dy/dx\).
Incorrect calculation: writing \( \sec^2 y=1\) after differentiating \( \tan y=x\) treats the derivative of \( \tan y\) as though \(y\) were the input itself. It does not account for how \(y\) changes when \(x\) changes.
Correction: because \(y=y(x)\), the chain rule gives
The correct equation is \( \sec^2 y\,\frac{dy}{dx}=1\). Solving it and then using \( \tan y=x\) produces \( \frac{dy}{dx}=\frac{1}{1+x^2}\). Keeping the dependent variable visible while differentiating is a reliable way to avoid this error.
Why the Derivation Works
Restrict tangent to \( (-\pi/2,\pi/2)\), where it is strictly increasing and takes every real value.
For \(y=\arctan x\), the relation is \( \tan y=x\).
The chain rule gives \( \sec^2 y\,\frac{dy}{dx}=1\).
Use \( \tan y=x\) and \( \sec^2 y=1+\tan^2 y\) to obtain \( \frac{dy}{dx}=\frac{1}{1+x^2}\).
The nonzero derivative of tangent is essential: it permits solving for \(dy/dx\). The range of the restricted tangent is also essential: it guarantees that the inverse, and hence the derivative calculation, applies for every real input. Together these facts explain both the formula and its full real domain.
Check Your Understanding
Use the defining relation \( \tan y=x\) and the derivative formula to answer these questions.
- What interval is used to restrict tangent when defining \( \arctan x\), and what is the range of the resulting inverse?
- Differentiate \( \tan y=x\) with respect to \(x\), remembering that \(y\) depends on \(x\).
- Use the identity \( \sec^2 y=1+\tan^2 y\) to derive the derivative formula for \( \arctan x\).
- Evaluate the derivative of \( \arctan x\) at \(x=4\), and show the arithmetic.
- Explain why \(1+x^2\) cannot be zero for any real \(x\).
