How do you find the derivative of #s=tsint#?

1 Answer
Mar 6, 2017

#s'(t)=sint+tcost#

Explanation:

This will require the product rule for derivatives.

Recall that the product rule states that given a function that is the product of two other functions,

#s(t)=f(t)*g(t)#

its derivative is

#s'(t)=f'(t)*g(t)+f(t)*g'(t)#

For this expression,

#f(t)=t#
and
#g(t)=sint#

So,

#s'(t)=(1)*sint+t*cost#

#s'(t)=sint+tcost#