Differentiate. \[ y=\left(r^{2}-6 r\right) e^{r} \] \( y^{\prime}=\square \)

To differentiate the function \( y = (r^{2} - 6r) e^{r} \), we can use the product rule. The product rule states that if you have two functions \( u(r) \) and \( v(r) \), then the derivative is given by \( (uv)' = u'v + uv' \). Here, let: - \( u = r^{2} - 6r \) - \( v = e^{r} \) Firstly, we find the derivative of \( u \): \[ u' = 2r - 6 \] Next, we find the derivative of \( v \): \[ v' = e^{r} \] Now, applying the product rule: \[ y' = u'v + uv' = (2r - 6)e^{r} + (r^{2} - 6r)e^{r} \] We can factor out \( e^{r} \): \[ y' = e^{r} \left( (2r - 6) + (r^{2} - 6r) \right) \] Combining like terms: \[ y' = e^{r} \left( r^{2} - 4r - 6 \right) \] So the derivative is: \[ y' = e^{r} (r^{2} - 4r - 6) \]