AP Calculus AB and BC

Derivative of arctan(sqrt x): Chain Rule

Q: What is the derivative of arctan(sqrt x)?

It is \frac{1}{2\sqrt{x}\left(1+x\right)}.

The derivative of arctan of the square root of x is 1 over 2 times the square root of x times 1 plus x. The arctangent derivative supplies 1 over 1 plus x, since the inner function squared is just x, and the chain rule multiplies by the square root derivative.

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

Why the denominator simplifies

\frac{d}{dx}\arctan u = \frac{u'}{1+u^{2}}, \quad u = \sqrt{x}, \quad u^{2} = x

Squaring the inner function removes the radical from that part entirely, which is why only one square root survives, in the chain rule factor.

\frac{d}{dx}\arctan\sqrt{x} = \frac{\frac{1}{2\sqrt{x}}}{1+x} = \frac{1}{2\sqrt{x}\left(1+x\right)}

Domain

The function needs $x \ge 0$ , and the derivative excludes $x = 0$ as well, where the tangent line is vertical.

Common mistakes

Writing $1 + x^{2}$ in the denominator. The inner function is $\sqrt{x}$ , so its square is $x$ .
Forgetting the chain rule factor $\frac{1}{2\sqrt{x}}$ .

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

What is the derivative of arctan(sqrt x)?

It is $\frac{1}{2\sqrt{x}\left(1+x\right)}$ .

Why is it 1 + x and not 1 + x^2?

Because the arctangent rule squares the INNER function, and $\left(\sqrt{x}\right)^{2} = x$ .