Differentiate with respect to t. y = b cos t + t2 sin t

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Differentiate with respect to t. y = b cos t + t2 sin t

Mathematics
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sin t (2-b) + 2t cos t
- sint
( cost)' ? (sint)' ?

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Other answers:

-> b ( cos t ) ' = - bsint Okie, so far?
okayy
y' = -bsin+2tcos
y=bcost+t2sint differentiating w.r.t t, dy/dt= -bsint+ d/dt(2tsint) now, we have to differentiate 2tsint using product rule for derivative. so, the derivative of 2tsint becomes (2tcost+2tsint) therefore, dy/dt= -bsint+2tsint+2tcost taking sint common, dy/dx=sint(2-b)+2tcost
i got it wrong
but i understand what you did
datz ok! :) not a big deal. derivatives is realy easy as compared to the other sections of calculus. u jus need 2 know the ways to differentiate! :)
yeaaa thank you :)
Apply product rule here ( t²* sin t ) ' with : u = t² --> u' = ... v = sint --> v' = ....
@v.s Do you know product rule?
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
y = b cos t + t² sin t -> b ( cos t ) ' = - bsint ( derivative of the first term) -> ( t² sin t )' = 2t sint+ t² cost ( derivative of the second term)
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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