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

How to differentiate e^x sin x

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

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

Both terms share a factor of $e^{x}$, so factor it out.

Why both terms add

Here neither derivative brings a minus sign: $e^{x}$ differentiates to $e^{x}$ and $\sin x$ differentiates to $+\cos x$. Contrast $e^{x}\cos x$, whose derivative is $e^{x}(\cos x - \sin x)$, where the minus sign comes from differentiating cosine.

The reverse operation is $\int e^{x}\sin x\,dx = \frac{e^{x}(\sin x - \cos x)}{2} + C$, found by integration by parts twice and solving for the original integral.

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

The product rule is Topic 2.8, on both AB and BC. Pairing $e^{x}$ with a trigonometric function is common, and the factored form $e^{x}(\sin x + \cos x)$ is the answer keys usually expect.

Common mistakes with the derivative of e^x sin x

• Answering $e^{x}\cos x$, multiplying the derivatives instead of using the product rule.
• Answering $e^{x}(\sin x - \cos x)$, adding a wrong minus sign; $\sin x$ differentiates to $+\cos x$.
• Answering $e^{x}(\cos x - \sin x)$, which is the derivative of $e^{x}\cos x$, not this one.