Derivative of sec x tan x: Proof and Common Mistakes

Product rule, then the identity

Both factors differentiate into something built from $\sec x$ and $\tan x$, which is why the raw answer looks bigger than it needs to. Take $u = \sec x$ with $u' = \sec x\tan x$, and $v = \tan x$ with $v' = \sec^{2}x$.

Now use the Pythagorean identity $\tan^{2}x = \sec^{2}x - 1$ on the first term. Everything becomes powers of $\sec x$, and a common factor of $\sec x$ falls out.

Both $\sec x\tan x$ and its derivative are undefined at the odd multiples of $\frac{\pi}{2}$, where $\cos x = 0$, so function and derivative share the same domain. Everything below, including the antiderivative statement, holds on each interval between consecutive asymptotes.

The derivative and the integral look nothing alike

$\sec x\tan x$ is one of the few expressions students meet far more often as an integrand than as a function to differentiate, because it is exactly the derivative of $\sec x$. Read backwards, that fact gives a one line antiderivative.

Going one step up in complexity and one step down are not mirror images here. The antiderivative is shorter than the input, the derivative is longer, and confusing the two directions is the most common way this expression goes wrong on a free response question.

The mistakes students make

Almost every wrong answer here comes from one of two places: the product rule itself, or the direction the Pythagorean identity is applied in.

• Stopping at $\sec x\tan^{2}x$. That is only the first product rule term; the second, $\sec x\cdot\sec^{2}x = \sec^{3}x$, is the one that gets dropped.
• Using the identity backwards as $\tan^{2}x = 1-\sec^{2}x$. That turns the sum into $\sec x - \sec^{3}x + \sec^{3}x = \sec x$, which is the antiderivative, not the derivative.
• Multiplying the two derivatives to get $\sec x\tan x\cdot\sec^{2}x = \sec^{3}x\tan x$. The product rule adds two terms; it never multiplies the derivatives together.