Question

can someone explain how from dy/dt=(2t+1)(3+cos(t^2)) you get yo d^2y/dt^2 @t=4=24.814

every time I try to do the second derivative and plug in 4, I get a different answer

Answer 1

what did u get before plugging in 4, as your second derivative ?

Answer 2

i got 2(3+cos(t^2)-2tsin(t^2)(2t+1)

Answer 3

the question is from a calc BC frq. I'm trying to do d atm

Answer 4

This is a product rule,
(fg)' = f'g+fg'
And when you do the derivative of the cosine
you have to use chain rule.