AP Calculus AB and BC

Derivative of arctan x: Answer, Proof, Mistakes

Q: Is the derivative of \arctan x the same as the derivative of \tan^{-1} x?

Yes. \arctan x and \tan^{-1} x are two notations for the same inverse tangent function, so both have derivative \frac{1}{1+x^2}. Watch the notation trap: here \tan^{-1} x means the inverse function, not (\tan x)^{-1} = \cot x, which is unrelated.

Q: Why does the trig disappear from the answer?

Implicit differentiation first gives \frac{1}{\sec^2 y}, which still hides the angle y. The identity \sec^2 y = 1 + \tan^2 y lets you substitute \tan y = x, so \sec^2 y = 1 + x^2. What is left is a rational function with no arctangent or secant in it.

Q: What is the domain of the derivative of \arctan x?

All real numbers. Since 1 + x^2 \geq 1 for every x, the denominator is never zero, so \frac{1}{1+x^2} is defined everywhere. That is different from \arcsin x, whose derivative \frac{1}{\sqrt{1-x^2}} only exists for |x| < 1.

Q: What is the second derivative of \arctan x?

Differentiate \frac{1}{1+x^2} = \left(1+x^2\right)^{-1} with the chain rule: \left(\left(1+x^2\right)^{-1}\right)' = -\left(1+x^2\right)^{-2} \cdot 2x = \dfrac{-2x}{\left(1+x^2\right)^2}. It is zero at x = 0, the inflection point of \arctan x.

The derivative of arctan x is 1/(1+x^2). It holds for every real number x, since arctangent is differentiable everywhere. For a composite arctan(u), the chain rule gives u'/(1+u^2). You get it by differentiating tan y = x implicitly and using the identity 1 + tan^2 y = sec^2 y.

\frac{d}{dx}\left[\arctan x\right] = \frac{1}{1+x^2}

The proof by implicit differentiation

Start from the definition of arctangent. If $y = \arctan x$ , then $\tan y = x$ , with $y$ restricted to $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ so that the inverse is single-valued.

Differentiate both sides with respect to $x$ . The left side needs the chain rule, since $y$ is itself a function of $x$ :

\sec^2 y \cdot \frac{dy}{dx} = 1

Solve for $\frac{dy}{dx}$ , then rewrite $\sec^2 y$ with the Pythagorean identity $1 + \tan^2 y = \sec^2 y$ :

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

Because $\tan y = x$ , replace $\tan^2 y$ with $x^2$ . Every trig piece cancels out and the result is a plain rational function:

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

Why no trig is left

The identity does the work. $\sec^2 y$ still hides an angle $y$ you cannot evaluate, but $1 + \tan^2 y$ trades it for $1 + x^2$ . That swap is why every inverse trig derivative comes out algebraic, with no arctangent or secant remaining.

The chain rule form, with two worked composites

On the AP exam the input is almost never a bare $x$ . Whenever arctangent wraps a function $u$ , multiply by that inner derivative:

\left(\arctan u\right)' = \dfrac{u'}{1 + u^2}

First composite: $u = 3x$ , so $u' = 3$ .

\left(\arctan(3x)\right)' = \dfrac{3}{1 + (3x)^2} = \dfrac{3}{1 + 9x^2}

Second composite: $u = x^2$ , so $u' = 2x$ .

\left(\arctan\left(x^2\right)\right)' = \dfrac{2x}{1 + \left(x^2\right)^2} = \dfrac{2x}{1 + x^4}

Square the whole input before adding 1: $(3x)^2 = 9x^2$ and $\left(x^2\right)^2 = x^4$ . Copying $u$ straight into the denominator without squaring it is the fastest way to lose the point.

Where it shows up on the AP exam

Inverse trig derivatives have their own place in the Course and Exam Description: Unit 3, Topic 3.4, Differentiating Inverse Trigonometric Functions. Unit 3 is worth 5-10% of both the AB and the BC exam. Arctangent's derivative is a clean rational function with no square root, which makes it a convenient inverse for both differentiation and antidifferentiation problems.

You will usually meet it inside a chain-rule composite (Topic 3.1) or folded into a product or quotient on the no-calculator section. The reverse direction matters just as much: since $\left(\arctan x\right)' = \dfrac{1}{1+x^2}$ , the antiderivative $\int \frac{1}{1+x^2}\,dx = \arctan x + C$ is a basic antiderivative rule in Unit 6, and completing the square (Topic 6.10) turns many integrals into exactly that arctangent form.

One fact, two directions

Learn $\left(\arctan x\right)' = \dfrac{1}{1+x^2}$ and you also own $\int \frac{1}{1+x^2}\,dx = \arctan x + C$ . Free-response scoring rewards students who spot a $1 + x^2$ in a denominator and reach for arctangent.

Common mistakes

Squaring the wrong thing. The denominator is $1 + x^2$ , a sum, not $(1 + x)^2$ . There is no cross term and nothing to expand.
Adding a stray negative sign. Arctangent's derivative is positive, $\frac{1}{1+x^2}$ . The minus sign belongs to its cousins: $\left(\operatorname{arccot} x\right)' = -\dfrac{1}{1+x^2}$ and $\left(\arccos x\right)' = -\dfrac{1}{\sqrt{1-x^2}}$ .
Confusing it with arcsine. $\left(\arcsin x\right)' = \dfrac{1}{\sqrt{1-x^2}}$ has a square root over a difference; arctangent has neither. Arctangent's derivative is defined for every real $x$ , while arcsine's needs $|x| < 1$ .
Dropping the chain rule. $\left(\arctan(5x)\right)' = \dfrac{5}{1+25x^2}$ , not $\dfrac{1}{1+25x^2}$ . The inner derivative multiplies the entire fraction.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

Is the derivative of $\arctan x$ the same as the derivative of $\tan^{-1} x$ ?

Yes. $\arctan x$ and $\tan^{-1} x$ are two notations for the same inverse tangent function, so both have derivative $\frac{1}{1+x^2}$ . Watch the notation trap: here $\tan^{-1} x$ means the inverse function, not $(\tan x)^{-1} = \cot x$ , which is unrelated.

Why does the trig disappear from the answer?

Implicit differentiation first gives $\frac{1}{\sec^2 y}$ , which still hides the angle $y$ . The identity $\sec^2 y = 1 + \tan^2 y$ lets you substitute $\tan y = x$ , so $\sec^2 y = 1 + x^2$ . What is left is a rational function with no arctangent or secant in it.

What is the domain of the derivative of $\arctan x$ ?

All real numbers. Since $1 + x^2 \geq 1$ for every $x$ , the denominator is never zero, so $\frac{1}{1+x^2}$ is defined everywhere. That is different from $\arcsin x$ , whose derivative $\frac{1}{\sqrt{1-x^2}}$ only exists for $|x| < 1$ .

What is the second derivative of $\arctan x$ ?

Differentiate $\frac{1}{1+x^2} = \left(1+x^2\right)^{-1}$ with the chain rule: $\left(\left(1+x^2\right)^{-1}\right)' = -\left(1+x^2\right)^{-2} \cdot 2x = \dfrac{-2x}{\left(1+x^2\right)^2}$ . It is zero at $x = 0$ , the inflection point of $\arctan x$ .