Where the derivative of arctan x comes from

By Marius, private math tutor in Montreal · 2026-10-05

The derivative of $\arctan x$ is $\frac{1}{1 + x^{2}}$ , for every real $x$ . It comes from writing $y = \arctan x$ as $\tan y = x$ , differentiating both sides, and reading one right triangle to put the answer back in terms of $x$ ; the same three steps give $\frac{1}{\sqrt{1 - x^{2}}}$ for $\arcsin x$ , so you can rebuild either formula on a test.

What arctan x means

$\arctan x$ is the angle whose tangent is $x$ , chosen strictly between $- \frac{π}{2}$ and $\frac{π}{2}$ :

y = \arctan x means \tan y = x .

That restriction is what makes the inverse a function, and the matching restriction for $\arcsin$ decides a sign further down. $\arctan x$ is also written $\tan^{- 1} x$ ; it is the same function.

The derivative, from tan y = x

Rewrite. $\tan y = x$ .
Differentiate both sides with respect to $x$ . The left side needs the chain rule, because $y$ depends on $x$ : $\sec^{2} y \frac{d y}{d x} = 1 .$
Solve. $\frac{d y}{d x} = \frac{1}{\sec^{2} y} = \cos^{2} y .$
Put the answer back in terms of $x$ . Step 3 is about the angle $y$ , and the question was about $x$ . This is the step the triangle is for.

The triangle

Draw a right triangle with an angle $y$ whose tangent is $x$ : opposite side $x$ , adjacent side $1$ . Pythagoras gives the hypotenuse, $\sqrt{1 + x^{2}}$ . Read the cosine off the triangle:

\cos y = \frac{1}{\sqrt{1 + x^{2}}},

so $\cos^{2} y = \frac{1}{1 + x^{2}}$ . Put that into step 3:

\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}} .

Each triangle is built from the equation it stands for: label two sides so that

\tan y = x

(left) or

\sin y = x

(right), then find the third with Pythagoras and read off

\cos y

A drawn triangle only shows $x > 0$ , an angle in the first quadrant. The identity $\sec^{2} y = 1 + \tan^{2} y$ covers every $x$ at once and agrees: $\sec^{2} y = 1 + x^{2}$ . Use whichever you can rebuild faster. And a quick sanity check: $1 + x^{2}$ is never below $1$ , so the derivative is always positive, which fits a graph that only ever rises.

Worked example: rebuild arctan(x/5) from scratch

The reason to know where the formula comes from is that the method works on functions you have not memorised. Take $f (x) = \arctan \frac{x}{5}$ and run the same steps.

Rewrite. $\tan y = \frac{x}{5}$ .
Differentiate. $\sec^{2} y \frac{d y}{d x} = \frac{1}{5}$ , so $\frac{d y}{d x} = \frac{\cos^{2} y}{5}$ .
Triangle. Opposite side $x$ , adjacent side $5$ , hypotenuse $\sqrt{25 + x^{2}}$ . So $\cos^{2} y = \frac{25}{25 + x^{2}}$ .
Substitute. $f^{'} (x) = \frac{1}{5} \cdot \frac{25}{25 + x^{2}} = \frac{5}{25 + x^{2}} .$

Check it with the chain rule and the formula just derived:

\frac{1}{1 + {(\frac{x}{5})}^{2}} \cdot \frac{1}{5} = \frac{25}{25 + x^{2}} \cdot \frac{1}{5} = \frac{5}{25 + x^{2}} .

The two routes agree. On a test, the second is faster once you trust the formula; the first is how you get the formula back when you do not.

The same three steps for arcsin

$y = \arcsin x$ means $\sin y = x$ with $- \frac{π}{2} \leq y \leq \frac{π}{2}$ .

Rewrite. $\sin y = x$ .
Differentiate. $\cos y \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} = \frac{1}{\cos y}$ .
Triangle (the right-hand one above). Opposite side $x$ , hypotenuse $1$ , so the adjacent side is $\sqrt{1 - x^{2}}$ , and $\cos y = \sqrt{1 - x^{2}}$ . For $- 1 < x < 1$ , $\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}} .$

Why the positive square root? Because $y$ lies in $[- \frac{π}{2}, \frac{π}{2}]$ , where the cosine is never negative. The restriction that made $\arcsin$ a function is the same one that fixes the sign. At $x = \pm 1$ that cosine is $0$ , which is why the formula stops short of the endpoints.

The common mistake: reading $\tan^{- 1} x$ as a reciprocal. Here the $- 1$ means the inverse function, $\arctan x$ , and the derivative is $\frac{1}{1 + x^{2}}$ . Read the $- 1$ as an exponent and you are differentiating $\frac{1}{\tan x} = \cot x$ instead, a different function whose derivative is $- \csc^{2} x$ . The derivative of $\arctan x$ has no trigonometric function left in it at all, so a $\csc$ or a $\cot$ in your answer is the sign that the $- 1$ was misread. The same goes for $\sin^{- 1} x$ , which means $\arcsin x$ .

Check your understanding

No calculator. Easiest first; questions 3, 6 and 7 ask for a reason as well as an answer.

Find the slope of $y = \arctan x$ at $x = \sqrt{3}$ .
Where is the graph of $y = \arctan x$ steepest, and what is its slope there?
Without differentiating, explain why the slope of $y = \arctan x$ must approach $0$ as $x \to \infty$ . Then confirm it from the formula.
Differentiate $g (x) = \arctan (e^{x})$ .
For which values of $x$ is the derivative of $\arcsin x$ equal to $2$ ?
$\arccos x$ takes its values in $[0, π]$ . Rebuild its derivative with the same three steps, starting from $\cos y = x$ , and explain your choice of sign for the square root.
Find the error: a student writes "Since $\tan y = x$ , $\frac{d y}{d x} = \frac{1}{\sec^{2} y}$ , so the derivative of $\arctan x$ is $\frac{1}{\sec^{2} x}$ ."
Calculus II: find a function whose derivative is $\frac{1}{4 + x^{2}}$ . The worked example is a good place to start.

Show the answers

$\frac{1}{4}$ , since $\frac{1}{1 + (\sqrt{3})^{2}} = \frac{1}{1 + 3}$ .
At $x = 0$ , with slope $1$ . The denominator $1 + x^{2}$ is smallest at $x = 0$ , so the derivative $\frac{1}{1 + x^{2}}$ is largest there.
The graph of $\arctan x$ is the graph of $\tan x$ on $(- \frac{π}{2}, \frac{π}{2})$ reflected in the line $y = x$ , which swaps rise and run. Near its vertical asymptote $x = \frac{π}{2}$ the graph of $\tan x$ gets steeper without bound, so its mirror image gets flatter and flatter as it levels off towards $y = \frac{π}{2}$ . (The asymptote alone would not settle it: a curve can approach a horizontal line while wiggling.) The formula agrees: $1 + x^{2}$ grows without bound, so $\frac{1}{1 + x^{2}} \to 0$ .
$g^{'} (x) = \frac{e^{x}}{1 + e^{2 x}}$ . The chain rule with $u = e^{x}$ gives $\frac{u^{'}}{1 + u^{2}}$ , and $u^{2} = e^{2 x}$ .
$x = \pm \frac{\sqrt{3}}{2}$ . Setting $\frac{1}{\sqrt{1 - x^{2}}} = 2$ gives $\sqrt{1 - x^{2}} = \frac{1}{2}$ , so $1 - x^{2} = \frac{1}{4}$ and $x^{2} = \frac{3}{4}$ ; both values lie inside $(- 1, 1)$ .
$\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}}$ for $- 1 < x < 1$ . Differentiating $\cos y = x$ gives $- \sin y \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} = - \frac{1}{\sin y}$ . The triangle (adjacent $x$ , hypotenuse $1$ ) gives $\sin y = \sqrt{1 - x^{2}}$ , with the positive root because $y$ lies in $[0, π]$ , where the sine is never negative.
The student replaced $y$ by $x$ at the end, but $y$ is the angle $\arctan x$ , not $x$ . $\sec^{2} y$ has to be rewritten in terms of $x$ : $\sec^{2} y = 1 + \tan^{2} y = 1 + x^{2}$ , which gives $\frac{1}{1 + x^{2}}$ . The student's version cannot be right: at $x = \frac{π}{2}$ , where $\sec x$ is undefined, it gives no slope at all, yet $\arctan x$ has a positive slope at every $x$ .
$\frac{1}{2} \arctan \frac{x}{2}$ , plus any constant. Check by the chain rule: $\frac{1}{2} \cdot \frac{1}{1 + x^{2} / 4} \cdot \frac{1}{2} = \frac{1}{4 + x^{2}}$ .

Practise it

The free CEGEP Calculus I Derivatives of Transcendental Functions worksheet has a sheet of inverse trigonometric derivatives, alongside exponentials, logarithms and logarithmic differentiation. The method in this post is implicit differentiation, which has its own sheet in the Differentiation Rules worksheet.

Want one-on-one help with this? I offer CEGEP Science calculus and linear algebra tutoring in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.