AP Calculus AB and BC

Derivative of x tan x: Answer, Proof, Mistakes

Q: What is the derivative of x tan x?

It is \tan x + x\sec^2 x. The product rule gives (1)\tan x + x\sec^2 x, using \frac{d}{dx}\tan x = \sec^2 x.

Q: What is the derivative of tan x?

It is \sec^2 x. From \tan x = \frac{\sin x}{\cos x}, the quotient rule and the identity \cos^2 x + \sin^2 x = 1 give \frac{1}{\cos^2 x} = \sec^2 x.

Q: Why are there two terms in the answer?

Because x\tan x is a product. The product rule differentiates each factor in turn and adds the results, producing \tan x from the first factor and x\sec^2 x from the second.

The derivative of x tan x with respect to x is tan x + x sec^2 x. The product rule differentiates x to 1, giving tan x, then keeps x and differentiates tan x to sec^2 x, giving x sec^2 x. Adding the two pieces yields tan x + x sec^2 x, valid wherever tan x is defined.

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

The proof: product rule on x tan x

This is a product of two functions, $f = x$ and $g = \tan x$ . The product rule is $f'g + fg'$ , with $f' = 1$ and $g' = \sec^2 x$ .

\frac{d}{dx}\left[x\tan x\right] = (1)\tan x + x(\sec^2 x) = \tan x + x\sec^2 x

The first term comes from differentiating $x$ and leaving $\tan x$ ; the second from keeping $x$ and differentiating $\tan x$ to $\sec^2 x$ . The key sub-fact is $\frac{d}{dx}\tan x = \sec^2 x$ .

Why the derivative of tan x is sec^2 x

Writing $\tan x = \frac{\sin x}{\cos x}$ and applying the quotient rule gives $\frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$ .

\frac{d}{dx}\tan x = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x

The Pythagorean identity $\cos^2 x + \sin^2 x = 1$ collapses the numerator to $1$ , and $\frac{1}{\cos^2 x}$ is $\sec^2 x$ . That is the $g'$ used above.

Common mistakes

Answering $\sec^2 x$ alone and forgetting the product rule entirely. The factor of $x$ means two terms, not one.
Writing $\tan x + \sec^2 x$ and dropping the $x$ on the second term. The product rule keeps $x$ attached to $\sec^2 x$ .
Using $\sec x\tan x$ for the derivative of $\tan x$ . That is the derivative of $\sec x$ ; the derivative of $\tan x$ is $\sec^2 x$ .
Multiplying the derivatives as $(1)(\sec^2 x)$ . The product rule adds two terms $f'g + fg'$ ; it does not multiply the two derivatives together.

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 x tan x?

It is $\tan x + x\sec^2 x$ . The product rule gives $(1)\tan x + x\sec^2 x$ , using $\frac{d}{dx}\tan x = \sec^2 x$ .

What is the derivative of tan x?

It is $\sec^2 x$ . From $\tan x = \frac{\sin x}{\cos x}$ , the quotient rule and the identity $\cos^2 x + \sin^2 x = 1$ give $\frac{1}{\cos^2 x} = \sec^2 x$ .

Why are there two terms in the answer?

Because $x\tan x$ is a product. The product rule differentiates each factor in turn and adds the results, producing $\tan x$ from the first factor and $x\sec^2 x$ from the second.