# Find the derivative of y = xe^-2x.

## Question:

Find the derivative of {eq}\ y = xe^{-2x} {/eq}.

## Product Rule:

Let {eq}f'(x) {/eq} and {eq}h'(x) {/eq} equal the derivatives of {eq}f(x) {/eq} and {eq}h(x) {/eq} respectively.

Adding {eq}f(x)h'(x) {/eq} and {eq}h(x)f'(x) {/eq} gives us the derivative of the product of {eq}f(x) {/eq} and {eq}h(x) {/eq} using the product rule.

## Answer and Explanation:

As {eq}y {/eq} consists of the product of two functions, we will employ the product rule to attain {eq}y' {/eq}:

{eq}\begin{align*} \displaystyle y& = xe^{-2x}\\ \displaystyle y'& = x\frac{\mathrm{d}}{\mathrm{d}x}(e^{-2x})+e^{-2x}\frac{\mathrm{d}}{\mathrm{d}x}(x) \ \ \ \left[\mathrm{Utilize \ the \ product \ rule }\right]\\ \displaystyle y'& = x (-2e^{-2x})+e^{-2x}(1)\\ \implies \displaystyle y'& = -2xe^{-2x}+e^{-2x}\\ \end{align*} {/eq}

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from AP Calculus AB & BC: Help and Review

Chapter 10 / Lesson 14