Answer
$$f_x = - \pi e^{-t}sin \pi x $$ $$f_t = - e^{-t}cos \pi x $$
Work Step by Step
$$f(x,t) = e^{-t}cos \pi x$$ $$u = \pi x, u' = \pi$$ $$f_x = (e^{-t})(-sin\pi x)(\pi)$$
$e^{-t}$ is constant.
(cos u)' = (- sin u) * u'
$$u = -t, u' = -1$$ $$f_t = -e^{-t}cos \pi x$$
$cos \pi x$ is constant.
$(e^u)' = (u')(e^{u})$
