## Calculus (3rd Edition)

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