Answer
$$ f'(t)
= - (1-2t)e^{-2t}\sin(te^{-2t})$$
Work Step by Step
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
$$ f'(t)= -\sin(te^{-2t}) (te^{-2t})'= - (e^{-2t}-2te^{-2t})\sin(te^{-2t})\\
= - (1-2t)e^{-2t}\sin(te^{-2t}).$$
