AP Calculus AB and BC

Derivative of sin 3x: Answer, Chain Rule, Mistakes

Q: What is the derivative of sin(kx)?

\frac{d}{dx}\sin(kx) = k\cos(kx). The chain rule multiplies \cos(kx) by the inner derivative k. With k = 3 that gives 3\cos 3x.

Q: Why is there a factor of 3?

Because the inside is 3x, whose derivative is 3. The chain rule requires multiplying by that inner derivative, so \cos 3x picks up a factor of 3.

Q: What is the second derivative of sin 3x?

Differentiate 3\cos 3x to get -9\sin 3x. Each derivative brings down another factor of 3 and cycles sine to cosine to negative sine.

The derivative of sin 3x with respect to x is 3 cos 3x. The chain rule differentiates the outer sine to cos 3x, then multiplies by the derivative of the inner function 3x, which is 3. The general pattern is that sin kx differentiates to k cos kx, so the factor of 3 is easy to miss but required.

\frac{d}{dx}\left[\sin 3x\right] = 3\cos 3x

How to differentiate sin 3x

The outer function is sine and the inner function is $3x$ . Differentiate the outer, leave the inner alone, then multiply by the inner derivative.

\frac{d}{dx}\sin(3x) = \cos(3x)\cdot 3 = 3\cos 3x

The same reasoning gives the general rule for any constant multiple inside the sine.

\frac{d}{dx}\sin(kx) = k\cos(kx)

Why the 3 appears

Multiplying the input by $3$ makes the argument sweep three times as fast, so the wave oscillates three times as quickly and its slopes are three times as steep. The factor of $3$ is that rate.

Reversed, the factor divides instead: $\int \sin 3x\,dx = -\frac{1}{3}\cos 3x + C$ . Forward it multiplies, backward it divides.

Where the derivative of sin 3x shows up on the AP exam

The chain rule is Topic 3.1, on both AB and BC, and a trigonometric function with a linear inner function is the standard first example. Unit 3 is weighted 5 to 10 percent on both exams.

Do not confuse it with the constant multiple $\frac{d}{dx}(3\sin x) = 3\cos x$ . Written quickly the two look alike, but they are different functions with different periods.

Common mistakes with the derivative of sin 3x

Answering $\cos 3x$ and dropping the factor of $3$ . This is the single most common chain rule slip.
Answering $3\cos x$ , which strips the $3$ out of the argument instead of leaving it inside the cosine.
Turning the sine into $-\cos$ . Sine differentiates to $+\cos$ ; the minus sign belongs to the derivative of cosine, not sine.

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 sin(kx)?

$\frac{d}{dx}\sin(kx) = k\cos(kx)$ . The chain rule multiplies $\cos(kx)$ by the inner derivative $k$ . With $k = 3$ that gives $3\cos 3x$ .

Why is there a factor of 3?

Because the inside is $3x$ , whose derivative is $3$ . The chain rule requires multiplying by that inner derivative, so $\cos 3x$ picks up a factor of $3$ .

What is the second derivative of sin 3x?

Differentiate $3\cos 3x$ to get $-9\sin 3x$ . Each derivative brings down another factor of $3$ and cycles sine to cosine to negative sine.