Derivative of e^2x sin x: Answer, Proof, Mistakes

How to differentiate e^(2x) sin x

This is a product of $u = e^{2x}$ and $v = \sin x$, so the product rule applies: $f' = u'v + uv'$. Finding $u'$ takes a chain rule step, because the exponent is $2x$ rather than $x$.

Both terms carry a factor of $e^{2x}$, so factoring gives the compact form graders expect.

Checking the answer and reading the graph

A fast check at $x = 0$: the formula gives $f'(0) = e^{0}(2\sin 0+\cos 0) = 1$, and the curve does leave the origin with slope $1$, since $\sin x \approx x$ there.

Because $e^{2x}$ is never zero, $f'(x) = 0$ only when $2\sin x+\cos x = 0$, that is when $\tan x = -\frac{1}{2}$. Those inputs are the turning points of an oscillation whose amplitude grows with $e^{2x}$.

On the AP exam this pairing is standard: the product rule is Topic 2.8 and the chain rule is Topic 3.1, and Unit 3 questions routinely nest one inside the other exactly like this.

Common mistakes with the derivative of e^(2x) sin x

• Multiplying the derivatives to get $2e^{2x}\cos x$. The product rule adds two terms, one for each factor differentiated in turn.
• Losing the chain rule factor and answering $e^{2x}(\sin x+\cos x)$. That is the pattern for $e^{x}\sin x$; the exponent $2x$ contributes a $2$ on the first term only.
• Factoring carelessly: $2e^{2x}\sin x+e^{2x}\cos x$ becomes $e^{2x}(2\sin x+\cos x)$, not $e^{2x}(2\sin x+2\cos x)$.