## Calculus: Early Transcendentals (2nd Edition)

$e^w(w^3+3w^2-1)$
We are trying to find $\frac{d}{dw}g(w)=e^w(w^3-1)$. Use the Product Rule, $\frac{d}{dw}u(w)v(w)=u'(w)v(w)+u(w)v'(w)$. In this case, $u(w)=e^w$, so $u'(w)=e^w$. Also, $v(w)=w^3-1$, so $v'(w)=3w^2$. Plugging these in, we get: $\frac{d}{dw}e^w(w^3-1)$ $=e^w(w^3-1)+e^w(3w^2)$ $=e^w(w^3+3w^2-1)$