Question

attempt 1: 10 attempts remaining. differentiate: ( y = (r^2 + 2r)e^r ) ( y = ) submit answer next item

Explanation:

Step1: Identify the product rule

We have \( y=(r^{2} + 2r)e^{r}\), which is a product of two functions \(u = r^{2}+2r\) and \(v = e^{r}\). The product rule for differentiation is \((uv)'=u'v + uv'\).

Step2: Differentiate \(u\) and \(v\)

First, find \(u'=\frac{d}{dr}(r^{2}+2r)=2r + 2\) (using the power rule \(\frac{d}{dr}(r^{n})=nr^{n - 1}\)).
Then, find \(v'=\frac{d}{dr}(e^{r})=e^{r}\) (since the derivative of \(e^{r}\) with respect to \(r\) is \(e^{r}\)).

Step3: Apply the product rule

Using \((uv)'=u'v+uv'\), we substitute \(u = r^{2}+2r\), \(u'=2r + 2\), \(v = e^{r}\), and \(v'=e^{r}\):
\[

$$\begin{align*} y'&=(2r + 2)e^{r}+(r^{2}+2r)e^{r}\\ &=e^{r}(2r + 2+r^{2}+2r)\\ &=e^{r}(r^{2}+4r + 2) \end{align*}$$

\]

Answer:

\(y'=(r^{2}+4r + 2)e^{r}\)