"fk378" <email@example.com> wrote in message news:33546210.1189959326672.JavaMail.firstname.lastname@example.org... > hello all! > > I'm having trouble with this problem: > Differentiate sin(t) + (pi)cos(t) > > So, I have f'(x) of sin(t) = cos(t) > Now what do I do with the (pi)cos(t) part? Do I say that the slope of pi > is zero, therefore the derivative of (pi)cos(t) is 0, then the answer > would be just sin(t) for the whole equation. > > OR > > Leave pi there, and have f'(x) of cos(t)= -sin(t) so that would make: > cos(t) + (pi)(-sin[t]) ??? > > thank you!!
The latter, which can be written cost(t)-(pi)sin(t). Pi is constant but pi(-sin(t)) is not, anymore than (1)(cos(t)) is even though 1 is constant.