Derivative of x^2 sin x: Answer, Product Rule, Mistakes

How to differentiate x^2 sin x

The expression is a product of $x^{2}$ and $\sin x$, so the product rule differentiates each factor once, holding the other fixed, and adds.

Take $f(x) = x^{2}$ and $g(x) = \sin x$, so $f'(x) = 2x$ and $g'(x) = \cos x$.

Unlike $x^{2}e^{x}$, the two terms share no common exponential or trigonometric factor to pull out. Factoring the common $x$ only gives $x(2\sin x + x\cos x)$, which is no simpler, so the expanded form $2x\sin x + x^{2}\cos x$ is the standard answer.

Why the product rule is necessary

If the derivative of a product were the product of the derivatives, the answer would be $2x\cos x$. Test it at $x = \frac{\pi}{2}$: the real derivative is $2\cdot\frac{\pi}{2}\cdot 1 + \left(\frac{\pi}{2}\right)^{2}\cdot 0 = \pi$, while $2x\cos x = \pi\cdot 0 = 0$. They disagree, so the shortcut is false.

The reverse operation, $\int x^{2}\sin x\,dx$, needs integration by parts applied twice.

Where the derivative of x^2 sin x shows up on the AP exam

The product rule is Topic 2.8, on both AB and BC. A polynomial times a trigonometric function is a standard example, because one factor differentiates to a simple power and the structure of the rule stays visible.

Common mistakes with the derivative of x^2 sin x

• Answering $2x\cos x$, multiplying the two derivatives instead of using the product rule.
• Answering $2x\sin x + \cos x$, dropping the $x^{2}$ that multiplies $\cos x$ in the second term.
• Answering $2x\sin x - x^{2}\cos x$, adding a wrong minus sign; $\sin x$ differentiates to $+\cos x$.