The derivative of arctan(x) is 1/(1+x²). Here's the proof so it actually sticks:

Let y = arctan(x), so tan(y) = x. Differentiating both sides implicitly:

$\sec^2(y) \cdot \frac{dy}{dx} = 1 \implies \frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y)$

From the right triangle where tan(y) = x: opposite = x, adjacent = 1, hypotenuse = √(1+x²), so cos(y) = 1/√(1+x²), giving cos²(y) = 1/(1+x²).

$\boxed{\frac{d}{dx}[\arctan(x)] = \frac{1}{1+x^2}}$

Worth memorising the full set: arcsin′(x) = 1/√(1−x²), arccos′(x) = −1/√(1−x²), arctan′(x) = 1/(1+x²).