Differentiate inverse tangent
A curve that flattens
For any real input \(x\), the curve \(y = \arctan x\) gives the angle \(y\) between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\) whose tangent is \(x\). The curve is green because it rises.
The angle stays between these values, so the curve comes close to the faint dashed lines at those heights but never reaches them.
At the orange point, \(x = 2\) and the tangent has slope \(\frac{1}{5}\), where 5 is \(1 + x^2 = 1 + 4\).
Drag the point or move the slider. As \(x\) grows, the curve flattens toward the upper dashed line, so the slope becomes smaller. The slope is largest at \(x = 0\), where it is 1.
One over the slope of the tangent curve
The figure now also shows the curve \(y = \tan x\). Because \(\tan y = x\), the green curve is \(y = \tan x\) with \(x\) and \(y\) swapped, which is its reflection in the faint diagonal \(y = x\). Press the button to see the reflection.
A reflection swaps the rise and the run of every tangent line. This means the slope of the green curve is 1 divided by the slope of \(y = \tan x\) at the matching orange point.
On \(y = \tan x\), the matching point has the angle \(y\) as its horizontal coordinate, so its slope is \(\sec^2 y\), which is at least 1. This means the slope of the green curve is positive and at most 1, and it exists for every \(x\).
Notation
The angle \(y\) is the inverse tangent of \(x\) when its tangent is \(x\), and \(y\) lies between \(-\frac{\pi}{2}\) and \(\frac{\pi}{2}\).
\[y = \arctan x \iff \tan y = x\]Differentiate both sides with respect to \(x\). The angle \(y\) depends on \(x\), so the chain rule gives \(\sec^2 y\) times \(\frac{dy}{dx}\).
\[\sec^2 y \cdot \frac{dy}{dx} = 1\]Divide by \(\sec^2 y\). This number is at least 1, so it is never 0.
\[\frac{dy}{dx} = \frac{1}{\sec^2 y}\]To write \(\sec^2 y\) in terms of \(x\), divide \(\sin^2 y + \cos^2 y = 1\) by \(\cos^2 y\). This gives the identity \(1 + \tan^2 y = \sec^2 y\).
\[\sec^2 y = 1 + \tan^2 y = 1 + x^2\]This derivative of the inverse tangent holds for every real \(x\).
\[\frac{d}{dx} \arctan x = \cy{\frac{1}{1 + x^2}}\]
When it fails
The denominator \(1 + x^2\) is at least 1, so this formula has no input where it fails.
At \(x = 4\), the slope is only \(\frac{1}{17}\), but it is positive because the curve still rises toward the dashed line. The inverse sine is different, because its derivative does not exist at \(x = 1\).
Example 1: a negative input
Find the slope of \(y = \arctan x\) at \(x = -3\).
Square the input. The square of a negative number is positive.
\[(-3)^2 = \cy{9}\]Substitute into the formula.
\[\frac{d}{dx} \arctan x \Big|_{x=-3} = \frac{1}{1 + 9} = \cy{\frac{1}{10}}\]The slope is positive although \(x\) is negative, because the curve rises everywhere. The small value matches the nearly flat tangent at \(x = -3\).
Example 2: no square root
Find the slope of \(y = \arctan x\) at \(x = 4\).
Square the input and add 1.
\[x^2 = 16, \qquad 1 + x^2 = \cy{17}\]A tempting wrong move: The formula for the inverse sine has a square root, so use a square root here.
\[\begin{aligned} \frac{d}{dx} \arctan x &= \cr{\frac{1}{\sqrt{1 + x^2}}} \\ &\approx \cr{0.24} \end{aligned}\]The red line is much steeper than the curve at the point. From \(x = 4\) to \(x = 5\), the curve rises only about 0.05, but a line with slope 0.24 rises about five times as much.
The formula for the inverse tangent has no square root, so divide 1 by \(1 + x^2\).
\[\frac{d}{dx} \arctan x \Big|_{x=4} = \cy{\frac{1}{17}}\]
Practice
Find the slope of \(y = \arctan x\) at \(x = \dfrac{1}{2}\).
Move the point to x = 1/2. The slope is 1 divided by 1 plus the square of x.
Solution
Find \(1 + x^2\).
\[1 + \frac{1}{4} = \cy{\frac{5}{4}}\]Divide 1 by this number. The slope is less than 1, so the curve is flatter here than at \(x = 0\).
\[\frac{1}{5/4} = \cy{\frac{4}{5}}\]
Questions
The figure shows \(y = \arctan x\), which rises, so it is green. Each question moves an orange tangent to the input it names, and the second question also shows faint tangents at \(x = 0\), 1 and 3.
What is the slope of \(y = \arctan x\) at \(x = 1.5\)?
- \(\tfrac{4}{13}\)
- \(\tfrac{2}{\sqrt{13}}\)
- \(\tfrac{2}{5}\)
- \(\tfrac{4}{9}\)
Answer
\(\tfrac{4}{13}\)
Where is the graph of \(y = \arctan x\) steepest?
- At \(x = 0\), where the slope is 1
- At \(x = 1\), where the slope is \(\tfrac{1}{2}\)
- At \(x = 3\), where the slope is \(\tfrac{1}{10}\)
- It is equally steep everywhere
Answer
At \(x = 0\), where the slope is 1
What is the slope of \(y = \arctan x\) at \(x = -\tfrac{1}{3}\)?
- \(\tfrac{9}{10}\)
- \(-\tfrac{9}{10}\)
- \(\tfrac{3}{2}\)
- \(\tfrac{3}{\sqrt{10}}\)
Answer
\(\tfrac{9}{10}\)
Summary
To differentiate \(\arctan x\), you divide 1 by \(1 + x^2\).
This works because the curve is \(y = \tan x\) reflected in the line \(y = x\), so its slope is \(\frac{1}{\sec^2 y} = \frac{1}{1 + x^2}\).
Because \(1 + x^2\) is at least 1, the formula holds for every real \(x\), and the slope is positive.
This is one of four skills in topic 3.4 of Which & Why: AP Calculus.