Differentiating Inverse Tangent

A2 · P3 · 9 min

The inverse tangent, $\tan^{-1}x$ , is the only inverse trigonometric function whose derivative is on the P3 syllabus. Its derivative, $\dfrac{1}{1 + x^2}$ , is a surprise: a function defined by angles has a derivative with no trigonometry in it at all. That fact is what makes it important. Reading it backwards gives the integral $\displaystyle\int\frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\frac{x}{a}$ , which appears in integration and differential equations questions. In differentiation, it turns up in composites, in products such as $x\tan^{-1}x$ , and in stationary-point questions.

What tan⁻¹ x means

$\tan x$ is not one-to-one, because it repeats every $\pi$ . To define an inverse, restrict $\tan$ to the interval $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ , where it takes every real value exactly once.

Definition

$y = \tan^{-1} x$ is the angle in the interval $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ whose tangent is $x$ . So $\tan^{-1} x = y \iff \tan y = x$ with $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ .

Its domain is all real $x$ and its range is $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ .

The graph is the reflection of the central branch of $y = \tan x$ in the line $y = x$ . The vertical asymptotes $x = \pm\tfrac{\pi}{2}$ of $\tan x$ become horizontal asymptotes $y = \pm\tfrac{\pi}{2}$ .

y = atan(x) y = pi/2 y = -pi/2

$y = \tan^{-1}x$ (solid) rises through the origin with gradient $1$ and flattens towards the asymptotes $y = \pm\tfrac{\pi}{2}$ (dashed).

From the graph you can predict the derivative: always positive, largest (equal to $1$ ) at $x = 0$ , decreasing towards zero as $|x|$ grows, and symmetric because the curve has rotational symmetry about the origin. $\dfrac{1}{1 + x^2}$ has exactly those features.

Watch out

$\tan^{-1} x$ is the inverse function, not the reciprocal. $\tan^{-1} x \ne \dfrac{1}{\tan x}$ . The reciprocal is $\cot x$ . Some books write $\arctan x$ for the inverse to avoid this confusion; Cambridge uses $\tan^{-1} x$ .

Deriving the derivative

Derivative of the inverse tangent

Let $y = \tan^{-1}x$ . Then $x = \tan y$ .

Differentiate with respect to $y$ :

\frac{dx}{dy} = \sec^2 y = 1 + \tan^2 y = 1 + x^2

using the identity $\sec^2 y = 1 + \tan^2 y$ and then $\tan y = x$ . Hence

\frac{dy}{dx} = \frac{1}{dx/dy} = \frac{1}{1 + x^2}

The crucial step is replacing $\sec^2 y$ with $1 + x^2$ . Leaving the answer as $\dfrac{1}{\sec^2 y}$ is correct but is in terms of $y$ , not $x$ ; the identity converts it.

The same proof can be done with implicit differentiation: differentiate $\tan y = x$ with respect to $x$ to get $\sec^2 y\dfrac{dy}{dx} = 1$ , and the rest is the same.

Derivative of tan⁻¹ x

\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2} \qquad\qquad \frac{d}{dx}\tan^{-1}f(x) = \frac{f'(x)}{1 + \big(f(x)\big)^2}

A useful special case:

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

That special case is the reason the integral of $\dfrac{1}{a^2 + x^2}$ is $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$ : see standard integrals.

Tip

The derivatives of $\sin^{-1}x$ and $\cos^{-1}x$ are not required by the 9709 syllabus. If you meet them elsewhere, the same method works: $y = \sin^{-1}x \Rightarrow x = \sin y \Rightarrow \dfrac{dx}{dy} = \cos y = \sqrt{1 - x^2}$ , so $\dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}}$ .

Composites, products and quotients

Everything you know about combining derivatives applies.

Function	Rule	Derivative
$\tan^{-1} 3x$	chain	$\dfrac{3}{1 + 9x^2}$
$\tan^{-1}(x^2)$	chain	$\dfrac{2x}{1 + x^4}$
$\tan^{-1}\sqrt{x}$	chain	$\dfrac{1}{2\sqrt{x}(1 + x)}$
$(\tan^{-1} x)^2$	chain (outer square)	$\dfrac{2\tan^{-1}x}{1 + x^2}$
$x\tan^{-1}x$	product	$\tan^{-1} x + \dfrac{x}{1 + x^2}$
$\tan^{-1}y$ with $y$ a function of $x$	implicit	$\dfrac{1}{1 + y^2}\dfrac{dy}{dx}$

Watch the square in the denominator: for $\tan^{-1} 3x$ it is $1 + (3x)^2 = 1 + 9x^2$ , not $1 + 3x^2$ .

The product $x\tan^{-1}x$ matters for another reason: its derivative contains $\tan^{-1}x$ , which is why the syllabus lists $x\tan^{-1}x$ as a standard integration by parts example.

Exact values you need

$x$	$0$	$\dfrac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	$-1$
$\tan^{-1} x$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$-\dfrac{\pi}{4}$

These come straight from the standard triangles. Note $\tan^{-1}(-x) = -\tan^{-1}x$ .

Worked examples

Composites

Differentiate (a) $\tan^{-1} 3x$ (b) $\tan^{-1}\dfrac{x}{2}$ (c) $\tan^{-1}(x^2)$ (d) $(\tan^{-1} x)^2$ .

Solution

(a) $\dfrac{1}{1 + (3x)^2}\times 3 = \dfrac{3}{1 + 9x^2}$ .

(b) $\dfrac{1}{1 + x^2/4}\times\dfrac{1}{2} = \dfrac{1}{2 + x^2/2}$ . Multiply top and bottom by $2$ : $\dfrac{2}{4 + x^2}$ . This matches $\dfrac{a}{a^2 + x^2}$ with $a = 2$ .

(d) Outer function is a square: $2\tan^{-1}x \times \dfrac{1}{1 + x^2} = \dfrac{2\tan^{-1}x}{1 + x^2}$ .

Product with the inverse tangent

Find the exact gradient of $y = x\tan^{-1}x$ at the point where $x = 1$ .

Solution

$u = x$ , $u' = 1$ ; $\ v = \tan^{-1}x$ , $v' = \dfrac{1}{1 + x^2}$ .

\frac{dy}{dx} = x\cdot\frac{1}{1 + x^2} + \tan^{-1}x

At $x = 1$ : $\dfrac{1}{2} + \tan^{-1}1 = \dfrac{1}{2} + \dfrac{\pi}{4}$ .

A tangent

Find the equation of the tangent to $y = \tan^{-1}(2x - 1)$ at the point where $x = 1$ .

Solution

At $x = 1$ : $y = \tan^{-1}1 = \tfrac{\pi}{4}$ .

$\dfrac{dy}{dx} = \dfrac{2}{1 + (2x - 1)^2}$ ; at $x = 1$ : $\dfrac{2}{1 + 1} = 1$ .

Tangent: $y - \tfrac{\pi}{4} = x - 1$ , i.e. $y = x - 1 + \tfrac{\pi}{4}$ .

A stationary point

Find the exact coordinates of the stationary point of $y = 2\tan^{-1}x - \ln(1 + x^2)$ and determine its nature.

Solution

\frac{dy}{dx} = \frac{2}{1 + x^2} - \frac{2x}{1 + x^2} = \frac{2(1 - x)}{1 + x^2}

Zero when $x = 1$ . Then $y = 2\cdot\tfrac{\pi}{4} - \ln 2 = \tfrac{\pi}{2} - \ln 2$ .

The denominator is positive, so the sign of $\dfrac{dy}{dx}$ is the sign of $1 - x$ : positive for $x < 1$ , negative for $x > 1$ . The point $\left(1,\ \tfrac{\pi}{2} - \ln 2\right)$ is a maximum.

Showing a function is always decreasing

Show that $f(x) = \tan^{-1}x - x$ is a decreasing function, and deduce that $\tan^{-1}x < x$ for all $x > 0$ .

Solution

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

This is negative for all $x \ne 0$ and zero only at the single point $x = 0$ , so $f$ is decreasing.

$f(0) = 0$ , and $f$ decreases, so for $x > 0$ : $f(x) < f(0) = 0$ , which means $\tan^{-1}x - x < 0$ , i.e. $\tan^{-1}x < x$ .

Exam-hard: a surprising derivative

Show that $\dfrac{d}{dx}\tan^{-1}\left(\dfrac{1 + x}{1 - x}\right) = \dfrac{1}{1 + x^2}$ for $x \ne 1$ , and explain this result using the identity for $\tan(A + B)$ .

Solution

Let $f(x) = \dfrac{1 + x}{1 - x}$ . By the quotient rule,

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

Also

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

\frac{d}{dx}\tan^{-1}f(x) = \frac{f'(x)}{1 + f(x)^2} = \frac{2}{(1-x)^2}\cdot\frac{(1 - x)^2}{2(1 + x^2)} = \frac{1}{1 + x^2}

Explanation: let $\theta = \tan^{-1}x$ . Then $\tan\left(\tfrac{\pi}{4} + \theta\right) = \dfrac{1 + \tan\theta}{1 - \tan\theta} = \dfrac{1 + x}{1 - x}$ . So for $x < 1$ , $\tan^{-1}\dfrac{1 + x}{1 - x} = \tfrac{\pi}{4} + \tan^{-1}x$ , which differs from $\tan^{-1}x$ by a constant and so has the same derivative.

Common mistakes

Forgetting to square the inside. $\dfrac{d}{dx}\tan^{-1}3x = \dfrac{3}{1 + 9x^2}$ , not $\dfrac{3}{1 + 3x^2}$ .
Forgetting the inner derivative on top. $\dfrac{d}{dx}\tan^{-1}(x^2) = \dfrac{2x}{1 + x^4}$ , not $\dfrac{1}{1 + x^4}$ .
Treating $\tan^{-1}$ as a reciprocal, giving $-\operatorname{cosec}^2 x$ (the derivative of $\cot x$ ).
Leaving the proof in terms of $y$ . In "show that $\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1+x^2}$ ", you must use $\sec^2 y = 1 + \tan^2 y = 1 + x^2$ explicitly.
Degrees. $\tan^{-1}1 = \tfrac{\pi}{4}$ , not $45$ , in any calculus answer.

Exam tip

The derivative of $\tan^{-1}x$ is in the list of formulae, but the composite forms are not. Expect to use the chain rule every time.
Exact answers involving $\tan^{-1}$ should use the standard values: $\tan^{-1}1 = \tfrac{\pi}{4}$ , $\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}$ .
When $\dfrac{dy}{dx}$ has a positive denominator like $1 + x^2$ , say so, then determine the sign of $\dfrac{dy}{dx}$ from the numerator alone. That is the cleanest route to the nature of a stationary point or to proving a function is increasing.
Expect $\tan^{-1}$ to resurface in the integration and differential equations parts of the paper: $\displaystyle\int\frac{1}{1 + x^2}\,dx = \tan^{-1}x + c$ .

Summary

$\tan^{-1}x$ is the angle in $\left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$ with tangent $x$ ; it is not $\cot x$ .
$\dfrac{d}{dx}\tan^{-1}x = \dfrac{1}{1 + x^2}$ , proved via $x = \tan y$ and $\sec^2 y = 1 + \tan^2 y$ .
Chain rule: $\dfrac{d}{dx}\tan^{-1}f(x) = \dfrac{f'(x)}{1 + f(x)^2}$ ; square the whole inside.
$\dfrac{d}{dx}\tan^{-1}\dfrac{x}{a} = \dfrac{a}{a^2 + x^2}$ , the source of $\displaystyle\int\frac{dx}{x^2 + a^2}$ .
Exact values: $\tan^{-1}1 = \tfrac{\pi}{4}$ , $\tan^{-1}\sqrt{3} = \tfrac{\pi}{3}$ , $\tan^{-1}\tfrac{1}{\sqrt{3}} = \tfrac{\pi}{6}$ .
$\sin^{-1}$ and $\cos^{-1}$ derivatives are not required.

Practice

Question

Differentiate $\tan^{-1}5x$ .
Differentiate $\tan^{-1}\dfrac{x}{\sqrt{3}}$ , giving your answer in the form $\dfrac{a}{b + x^2}$ .
Differentiate $\tan^{-1}\sqrt{x}$ .
Differentiate $x^2\tan^{-1}x$ .
Show that $\dfrac{d}{dx}\tan^{-1}\left(\dfrac{1}{x}\right) = -\dfrac{1}{1 + x^2}$ for $x \neq 0$ . What does this tell you about $\tan^{-1}x + \tan^{-1}\dfrac{1}{x}$ for $x > 0$ ?
Find the equation of the tangent to $y = \tan^{-1}x$ at the point where $x = \sqrt{3}$ .
Find the exact coordinates of the stationary points of $y = x - 2\tan^{-1}x$ and determine their nature.
Find the equation of the tangent to $y = \tan^{-1}(e^{2x})$ at the point where it crosses the $y$ -axis.
Show that $y = \tan^{-1}x - \dfrac{x}{1 + x^2}$ is an increasing function, and identify the nature of its stationary point.

Answers

$\dfrac{5}{1 + 25x^2}$ .
$\dfrac{1/\sqrt{3}}{1 + x^2/3} = \dfrac{\sqrt{3}}{3 + x^2}$ .
$\dfrac{1}{1 + x}\cdot\dfrac{1}{2\sqrt{x}} = \dfrac{1}{2\sqrt{x}(1 + x)}$ .
$2x\tan^{-1}x + \dfrac{x^2}{1 + x^2}$ .
$\dfrac{1}{1 + 1/x^2}\cdot\left(-\dfrac{1}{x^2}\right) = -\dfrac{1}{x^2 + 1}$ . So $\tan^{-1}x + \tan^{-1}\dfrac{1}{x}$ has derivative $0$ and is constant for $x > 0$ . At $x = 1$ it equals $\tfrac{\pi}{4} + \tfrac{\pi}{4} = \tfrac{\pi}{2}$ , so it is $\tfrac{\pi}{2}$ for all $x > 0$ (complementary angles in a right-angled triangle).
$y = \tfrac{\pi}{3}$ , gradient $\dfrac{1}{1 + 3} = \dfrac{1}{4}$ . Tangent: $y - \tfrac{\pi}{3} = \tfrac{1}{4}(x - \sqrt{3})$ .
$\dfrac{dy}{dx} = 1 - \dfrac{2}{1 + x^2} = \dfrac{x^2 - 1}{1 + x^2} = 0 \Rightarrow x = \pm 1$ . At $x = 1$ : $y = 1 - \tfrac{\pi}{2}$ ; at $x = -1$ : $y = -1 + \tfrac{\pi}{2}$ . $\dfrac{d^2y}{dx^2} = \dfrac{4x}{(1 + x^2)^2}$ : positive at $x = 1$ (minimum at $\left(1, 1 - \tfrac{\pi}{2}\right)$ ), negative at $x = -1$ (maximum at $\left(-1, \tfrac{\pi}{2} - 1\right)$ ).
At $x = 0$ : $y = \tan^{-1}1 = \tfrac{\pi}{4}$ . $\dfrac{dy}{dx} = \dfrac{2e^{2x}}{1 + e^{4x}}$ , which is $\dfrac{2}{2} = 1$ at $x = 0$ . Tangent: $y = x + \tfrac{\pi}{4}$ .
$\dfrac{dy}{dx} = \dfrac{1}{1 + x^2} - \dfrac{(1 + x^2) - x\cdot 2x}{(1 + x^2)^2} = \dfrac{(1 + x^2) - (1 - x^2)}{(1 + x^2)^2} = \dfrac{2x^2}{(1 + x^2)^2}$ . This is $\ge 0$ for all $x$ and zero only at $x = 0$ , so $y$ is increasing. At $x = 0$ the gradient is positive on both sides, so the stationary point $(0, 0)$ is a point of inflexion.

What tan⁻¹ x means

Deriving the derivative

Composites, products and quotients

Exact values you need

Worked examples

Practice

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