Calculus (3rd Edition)

Since we have $$y= e^{t-t^2}$$ then the first derivative $\frac{dy}{dt}$, using the chain rule, is given by \begin{align*} \frac{dy}{dt}&= e^{t-t^2} (t-t^2)'\\ &=(1-2t)e^{t-t^2} \end{align*} The second derivative $\frac{d^2y}{dt^2}$, using the product rule, is given by \begin{align*}\frac{d^2y}{dt^2}&=(1-2t)'e^{t-t^2}+(1-2t)(e^{t-t^2})' \\ &=(-2)e^{t-t^2}+(1-2t)(1-2t)e^{t-t^2}\\ &= \left(4 t^{2}-4 t-1\right) e^{t-t^{2}}. \end{align*}