## Calculus (3rd Edition)

$$g'(t)= (2t-1)e^{1/t}.$$
Recall that $(e^x)'=e^x$ Recall the product rule: $(uv)'=u'v+uv'$ Since $g(t)=t^2e^{1/t}$, then the derivative, using the product and chain rules, is given by $$g'(t)=(t^2)'e^{1/t}+t^2(e^{1/t})'=2t e^{1/t}+t^2e^{1/t} (\frac{-1}{t^2})=(2t-1)e^{1/t}.$$