The Curve Whose Area Hides π
Chase the area under the curve $y = \dfrac{1}{1+x^2}$ far enough and the number $\pi$ falls out of it. That gentle, bell-shaped curve is the derivative of tan inverse x, and because integrating it rebuilds $\tan^{-1}x$, it became one of the first reliable machines for computing $\pi$ to hundreds of digits. The whole chain rests on knowing what $\tan^{-1}x$ differentiates to.
That reach is why the result earns its keep well past calculus class: the same $\frac{1}{1+x^2}$ shape is the Cauchy–Lorentz curve in physics and the derivative used inside an arctangent activation when a neural network learns.
What Is the Derivative of Tan Inverse x?
The derivative of tan inverse x is $\dfrac{1}{1+x^2}$. In full notation:
$$\frac{d}{dx}\big(\tan^{-1}x\big) = \frac{1}{1+x^2}$$
Here $\tan^{-1}x$ (also written $\arctan x$) is the inverse tangent — the function that takes a ratio and returns the angle whose tangent is that ratio, with its principal value restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. The derivative gives the slope of that S-shaped arctan curve at each input $x$, and it belongs to the family of inverse trigonometric functions derivatives.
Two facts about the answer are worth fixing before any proof. First, the derivative is always positive and never larger than $1$: arctangent only ever rises, most steeply at $x = 0$ where the slope equals $1$, and it flattens toward the horizontal asymptotes. Second, unlike most trig derivatives it is a rational function with no trig left in it, which is why $\frac{1}{1+x^2}$ shows up so often on the integration side of the table in differentiation of trigonometric functions.
How Do You Prove the Derivative of Tan Inverse x?
The standard proof turns the inverse relationship into an ordinary tangent equation, then differentiates implicitly. The first principle confirms it from the limit definition.
Proof by Implicit Differentiation
Let $y = \tan^{-1}x$. By the definition of inverse tangent, this is the same statement as:
$$\tan y = x$$
Differentiate both sides with respect to $x$. The left side needs the chain rule, since $y$ depends on $x$:
$$\sec^2 y ,\frac{dy}{dx} = 1$$
Solve for $\dfrac{dy}{dx}$:
$$\frac{dy}{dx} = \frac{1}{\sec^2 y}$$
Now convert $\sec^2 y$ back into $x$ using the Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ from the trigonometric identities. Since $\tan y = x$, we have $\tan^2 y = x^2$, so:
$$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$
That substitution, trading $\sec^2 y$ for $1 + x^2$, is the move that clears every trace of $y$ out of the answer.
Proof by the First Principle
The limit definition confirms the shortcut is honest. With $f(x) = \tan^{-1}x$:
$$f'(x) = \lim_{h \to 0} \frac{\tan^{-1}(x+h) - \tan^{-1}x}{h}$$
Use the subtraction identity $\tan^{-1}A - \tan^{-1}B = \tan^{-1}\left(\dfrac{A - B}{1 + AB}\right)$ with $A = x+h$ and $B = x$:
$$f'(x) = \lim_{h \to 0} \frac{1}{h},\tan^{-1}\left(\frac{h}{1 + x(x+h)}\right)$$
As $h \to 0$ the argument of $\tan^{-1}$ becomes small, and for small angles $\tan^{-1}\theta \approx \theta$, so the arctangent and its argument cancel in the limit:
$$f'(x) = \lim_{h \to 0} \frac{1}{h} \times \frac{h}{1 + x(x+h)} = \frac{1}{1 + x^2}$$
Same destination, no rule assumed. A companion relationship worth noting: differentiating $\tan^{-1}x$ with respect to $\cot^{-1}x$ gives $-1$, because their derivatives are exact negatives — the same pairing that governs the inverse trigonometric ratios.
Examples of the Derivative of Tan Inverse x
Example 1
Differentiate $y = \tan^{-1}x$ and state the slope at $x = 0$.
$$\frac{dy}{dx} = \frac{1}{1+x^2}$$
At $x = 0$, the slope is $\dfrac{1}{1 + 0} = 1$, the steepest the arctangent curve ever gets.
Final answer: $\dfrac{dy}{dx} = \dfrac{1}{1+x^2}$. The slope at the origin is $1$.
Example 2
Differentiate $y = \tan^{-1}(x^2)$. First instinct, then the correct route.
The tempting move is to copy the base rule and write $\dfrac{dy}{dx} = \dfrac{1}{1 + x^2}$, swapping $x$ for $x^2$ only inside the visible slot and stopping there.
Check what changed. The input is now $x^2$, a function of $x$, so the chain rule demands a factor for its derivative. Writing $\dfrac{1}{1 + (x^2)^2}$ alone ignores how fast the inside is moving, and the answer comes out too small everywhere except $x = 0$.
The rescue multiplies by the inner derivative $\dfrac{d}{dx}(x^2) = 2x$:
$$\frac{dy}{dx} = \frac{1}{1 + (x^2)^2} \times 2x = \frac{2x}{1 + x^4}$$
Final answer: $\dfrac{dy}{dx} = \dfrac{2x}{1 + x^4}$.
Example 3
Differentiate $y = \tan^{-1}(3x)$.
Chain rule with inner derivative $\dfrac{d}{dx}(3x) = 3$:
$$\frac{dy}{dx} = \frac{1}{1 + (3x)^2} \times 3 = \frac{3}{1 + 9x^2}$$
Final answer: $\dfrac{dy}{dx} = \dfrac{3}{1 + 9x^2}$.
Example 4
Evaluate the derivative of $\tan^{-1}x$ at $x = 1$.
$$\left.\frac{d}{dx}\big(\tan^{-1}x\big)\right|_{x=1} = \frac{1}{1 + 1^2} = \frac{1}{2}$$
Final answer: the slope at $x = 1$ is $\dfrac{1}{2}$.
Example 5
Differentiate $y = x,\tan^{-1}x$.
Product rule, with $\dfrac{d}{dx}(x) = 1$ and $\dfrac{d}{dx}\big(\tan^{-1}x\big) = \dfrac{1}{1+x^2}$:
$$\frac{dy}{dx} = (1)\tan^{-1}x + x \times \frac{1}{1+x^2} = \tan^{-1}x + \frac{x}{1+x^2}$$
Final answer: $\dfrac{dy}{dx} = \tan^{-1}x + \dfrac{x}{1+x^2}$.
Example 6
Differentiate $y = \tan^{-1}\left(\dfrac{1}{x}\right)$.
Chain rule with inner derivative $\dfrac{d}{dx}\left(\dfrac{1}{x}\right) = -\dfrac{1}{x^2}$:
$$\frac{dy}{dx} = \frac{1}{1 + \left(\frac{1}{x}\right)^2} \times \left(-\frac{1}{x^2}\right)$$
Simplify the first factor by multiplying top and bottom by $x^2$:
$$= \frac{x^2}{x^2 + 1} \times \left(-\frac{1}{x^2}\right) = -\frac{1}{1 + x^2}$$
Final answer: $\dfrac{dy}{dx} = -\dfrac{1}{1 + x^2}$, the exact negative of the base derivative, because $\tan^{-1}\frac{1}{x} = \frac{\pi}{2} - \tan^{-1}x$ for positive $x$.
The first-instinct error across these is treating the argument as if it were a bare $x$: the moment the input is $x^2$, $3x$, or $\frac{1}{x}$, the chain-rule factor is what keeps the rate honest.
Why Does the Derivative of Tan Inverse x Matter?
The derivative $\frac{1}{1+x^2}$ is quietly one of the most reused expressions in mathematics, because it is what you integrate to get an angle back — the reverse direction is where its real work happens.
Computing π. Since $\tan^{-1}1 = \frac{\pi}{4}$, integrating $\frac{1}{1+x^2}$ term by term produces the Leibniz formula for π, $\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots$, and its faster Machin-like cousins ran the first multi-hundred-digit π calculations.
Physics and statistics. The same $\frac{1}{1+x^2}$ shape is the unnormalised Cauchy–Lorentz distribution, which models resonance line-widths and heavy-tailed data.
Machine learning. An arctangent activation function uses $\frac{1}{1+x^2}$ as its gradient during backpropagation, keeping updates smooth and bounded.
What competitor explainers rarely say plainly is why the slope maxes out at $x = 0$: arctangent climbs fastest where tangent is gentlest, near the origin, and it flattens toward $\pm\frac{\pi}{2}$ because tangent races to infinity there, so equal ratio-changes buy smaller and smaller angle-changes. The derivative reads that flattening directly.
Common Mistakes With the Derivative of Tan Inverse x
Mistake 1: Forgetting the Chain-Rule Factor
Where it slips in: Whenever the argument is $\tan^{-1}(kx)$, $\tan^{-1}(x^2)$, or any inside function other than a bare $x$.
Don't do this: Writing $\dfrac{d}{dx}\tan^{-1}(5x) = \dfrac{1}{1 + 25x^2}$ and stopping.
The correct way: Multiply by the derivative of the inside: $\dfrac{5}{1 + 25x^2}$. The rusher plugs the inside into the formula and moves on; the habit the chain rule installs is to write the inner derivative as a separate factor every time before simplifying.
Mistake 2: Confusing the Inverse With the Reciprocal
Where it slips in: Reading $\tan^{-1}x$ quickly, where the $-1$ superscript looks like a power.
Don't do this: Treating $\tan^{-1}x$ as $\dfrac{1}{\tan x} = \cot x$ and differentiating that instead.
The correct way: $\tan^{-1}x$ is the inverse function arctangent, not the reciprocal $\frac{1}{\tan x}$. The point of confusion is that the same $-1$ notation means "inverse function" on $\tan^{-1}x$ but "reciprocal" on $x^{-1}$. This is the identical trap that catches the derivative of cosec x, so fixing the reading habit once pays off across every inverse-trig derivative.
Mistake 3: Mishandling the Sign on $\tan^{-1}(1/x)$ Style Arguments
Where it slips in: Composite arguments whose own derivative is negative, like $\frac{1}{x}$ or $\cos x$.
Don't do this: Writing $\dfrac{d}{dx}\tan^{-1}\left(\frac{1}{x}\right) = \dfrac{1}{1 + \frac{1}{x^2}}$ and dropping the negative from the inner derivative.
The correct way: The inner derivative of $\frac{1}{x}$ is $-\frac{1}{x^2}$, and that minus sign survives all the way to the final answer $-\frac{1}{1+x^2}$ (Example 6). The memorizer who recalls only the outer formula loses the sign, while carrying the inner derivative as an explicit factor keeps it.
Key Takeaways
The derivative of tan inverse x is $\dfrac{d}{dx}\big(\tan^{-1}x\big) = \dfrac{1}{1+x^2}$, positive for all real $x$.
The implicit-differentiation proof rewrites $y = \tan^{-1}x$ as $\tan y = x$, then swaps $\sec^2 y$ for $1 + x^2$.
The slope peaks at $1$ when $x = 0$ and flattens toward zero as $x \to \pm\infty$.
For a composite argument, multiply by the inner derivative: $\dfrac{d}{dx}\tan^{-1}(kx) = \dfrac{k}{1 + k^2x^2}$.
Arctangent is the inverse function, not the reciprocal $\frac{1}{\tan x}$, and the two differentiate to completely different results.
To take these inverse-trig derivatives further with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for calculus prep, or live math tutoring with peers from 20+ countries.
A Practical Next Step
Practice these to solidify your understanding: differentiate $\tan^{-1}(4x)$, then $x^2\tan^{-1}x$, writing the inner derivative as its own factor each time. If you get stuck on a composite argument, come back to Example 2 above. Want a live Bhanzu trainer to work through inverse-trig differentiation with you? Book a free demo class.
Read More
Derivative of Arccos — the inverse-cosine derivative, proved the same implicit way.
Sin Cos Tan — the primary ratios that inverse-trig functions undo.
Arctan 1 — the value $\tan^{-1}1 = \frac{\pi}{4}$ used in the π connection.
Derivative of Tan 2x — a chain-rule derivative of a composite tangent function.
Trigonometric Table — the special-angle values behind these derivatives.
Was this article helpful?
Your feedback helps us write better content