AP Calculus AB and BC

Derivative of sec^3 x: Answer, Proof, and Mistakes

The derivative of sec^3 x is 3 sec^3 x tan x. Treat it as a power of sec x and use the chain rule: the power rule drops one secant and the inner derivative sec x tan x puts one straight back, so three factors of secant survive in the answer.

ddx[sec3x]=3sec3xtanx\frac{d}{dx}\left[\sec^{3}x\right] = 3\sec^{3}x\tan x

Chain rule with the power on the outside

Read sec3x\sec^{3}x as (secx)3\left(\sec x\right)^{3}. The outer function is the cube, the inner function is secx\sec x, and the inner derivative is secxtanx\sec x\tan x.

ddx(secx)3=3(secx)2secxtanx\frac{d}{dx}\left(\sec x\right)^{3} = 3\left(\sec x\right)^{2}\cdot\sec x\tan x

Now count secants. The power rule hands back sec2x\sec^{2}x, one power lower than you started with, and the inner derivative contributes one more. Two plus one is three, so the exponent returns to where it began.

f(x)=3sec3xtanxf'(x) = 3\sec^{3}x\tan x

Domain, shape and the integral that comes free

secx\sec x is undefined wherever cosx=0\cos x = 0, so both ff and ff' live on intervals such as (π2,π2)\left(-\frac{\pi}{2},\frac{\pi}{2}\right). The sample inputs x=0.3,0.7,1.1x = 0.3, 0.7, 1.1 all sit inside that interval.

On that interval sec3x\sec^{3}x is positive, so the sign of ff' is the sign of tanx\tan x. The function falls on (π2,0)\left(-\frac{\pi}{2},0\right) and rises on (0,π2)\left(0,\frac{\pi}{2}\right), with a minimum at x=0x=0 of value sec30=1\sec^{3}0 = 1.

Read the result backwards and it is an antiderivative in disguise: since ddx(13sec3x)=sec3xtanx\frac{d}{dx}\left(\frac{1}{3}\sec^{3}x\right) = \sec^{3}x\tan x, you get sec3xtanxdx=13sec3x+C\int \sec^{3}x\tan x\,dx = \frac{1}{3}\sec^{3}x+C without any work, which is the u=secxu=\sec x substitution.

The mistakes students make

The first two wrong answers come from mishandling the inner derivative of sec x, and the third comes from misreading the notation.

  • Answering 3sec2xtanx3\sec^{2}x\tan x. That is the power rule multiplied by tanx\tan x alone. The inner derivative is secxtanx\sec x\tan x, and the extra secx\sec x pushes the power back up to 33.
  • Answering 3sec2x3\sec^{2}x, treating secx\sec x as if it were the variable and stopping after the power rule. The chain rule is not optional once the inner function is anything other than xx.
  • Reading the function as sec(x3)\sec\left(x^{3}\right) and answering 3x2sec(x3)tan(x3)3x^{2}\sec\left(x^{3}\right)\tan\left(x^{3}\right). The notation sec3x\sec^{3}x cubes the output, not the input.

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 sec^3 x?

It is 3sec3xtanx3\sec^{3}x\tan x. The chain rule gives 3sec2xsecxtanx3\sec^{2}x\cdot\sec x\tan x, and the two secant powers combine to sec3x\sec^{3}x.

Why does the answer keep sec^3 instead of sec^2?

The power rule lowers the exponent to 22, but the inner derivative secxtanx\sec x\tan x supplies another factor of secx\sec x. Multiplying sec2x\sec^{2}x by secx\sec x returns the exponent to 33.

Is sec^3 x the same as sec(x^3)?

No. sec3x\sec^{3}x means (secx)3\left(\sec x\right)^{3}, so the cube is applied last. Its derivative is 3sec3xtanx3\sec^{3}x\tan x, while the derivative of sec(x3)\sec\left(x^{3}\right) is 3x2sec(x3)tan(x3)3x^{2}\sec\left(x^{3}\right)\tan\left(x^{3}\right).