AP Calculus AB and BC

Derivative of sqrt(tan x): Chain Rule

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

It is \frac{\sec^{2}x}{2\sqrt{\tan x}}.

Q: What is the domain?

Where \tan x \ge 0, such as \left[0, \frac{\pi}{2}\right), repeating with period \pi. The derivative additionally excludes the points where \tan x = 0.

The derivative of the square root of tan x is secant squared x divided by twice the square root of tan x. The power rule handles the outer root and the chain rule multiplies by secant squared x, the derivative of the inside.

\frac{d}{dx}\left[\sqrt{\tan x}\right] = \frac{\sec^{2}x}{2\sqrt{\tan x}}

Power rule over chain rule

\frac{d}{dx}(\tan x)^{1/2} = \frac{1}{2}(\tan x)^{-1/2}\cdot\sec^{2}x = \frac{\sec^{2}x}{2\sqrt{\tan x}}

The domain is the interesting part

The square root needs $\tan x \ge 0$ , which holds on $\left[0, \frac{\pi}{2}\right)$ and repeats with period $\pi$ . Where tangent is negative the function does not exist at all.

At $x = 0$ the function is $0$ but the derivative is undefined, because the denominator vanishes: a vertical tangent, just as with $\sqrt{x}$ at the origin.

Common mistakes

Answering $\frac{\sec^{2}x}{2}$ , forgetting that the power rule leaves the inside under the root.
Ignoring the domain and evaluating where $\tan x < 0$ .
Using $\sec x\tan x$ as the inner derivative. That belongs to secant.

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 sqrt(tan x)?

It is $\frac{\sec^{2}x}{2\sqrt{\tan x}}$ .

What is the domain?

Where $\tan x \ge 0$ , such as $\left[0, \frac{\pi}{2}\right)$ , repeating with period $\pi$ . The derivative additionally excludes the points where $\tan x = 0$ .