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sin t (2-b) + 2t cos t

- sint

( cost)' ? (sint)' ?

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-> b ( cos t ) ' = - bsint
Okie, so far?

okayy

y' = -bsin+2tcos

i got it wrong

but i understand what you did

yeaaa thank you :)

Apply product rule here ( t²* sin t ) ' with :
u = t² --> u' = ...
v = sint --> v' = ....

yes

2t*sint + cost*t^2

Yep, now put the 2 results together :)
y' = ....

y = b cos t + t² sin t
->y' = ...

y"= -bcost+2t*sint+cost*t^2

y = b cos t + t² sin t
-> b ( cos t ) ' = - bsint
-> ( t² sin t )' = 2t sint+ t² cost

you just put them together

- bsint + 2t sint+ t² cost

What's my instruction above?

product rule?

The product rule applied to take derivative of the second term!

i'm confused

what do you do next?

Yep, now put the 2 results together :)
y = b cos t + t² sin t
y' = - bsint + 2t sint+ t² cost

Now you can make it neat by simplifying - bsint + 2t sint together.

okaay
thnks you

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