anonymous
  • anonymous
Differentiate with respect to t. y = b cos t + t2 sin t
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
sin t (2-b) + 2t cos t
anonymous
  • anonymous
- sint
anonymous
  • anonymous
( cost)' ? (sint)' ?

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anonymous
  • anonymous
-> b ( cos t ) ' = - bsint Okie, so far?
anonymous
  • anonymous
okayy
anonymous
  • anonymous
y' = -bsin+2tcos
anonymous
  • anonymous
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
anonymous
  • anonymous
i got it wrong
anonymous
  • anonymous
but i understand what you did
anonymous
  • anonymous
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! :)
anonymous
  • anonymous
yeaaa thank you :)
anonymous
  • anonymous
Apply product rule here ( t²* sin t ) ' with : u = t² --> u' = ... v = sint --> v' = ....
anonymous
  • anonymous
@v.s Do you know product rule?
anonymous
  • anonymous
yes
anonymous
  • anonymous
2t*sint + cost*t^2
anonymous
  • anonymous
Yep, now put the 2 results together :) y' = ....
anonymous
  • anonymous
y = b cos t + t² sin t ->y' = ...
anonymous
  • anonymous
y"= -bcost+2t*sint+cost*t^2
anonymous
  • anonymous
y = b cos t + t² sin t -> b ( cos t ) ' = - bsint -> ( t² sin t )' = 2t sint+ t² cost
anonymous
  • anonymous
you just put them together
anonymous
  • anonymous
- bsint + 2t sint+ t² cost
anonymous
  • anonymous
What's my instruction above?
anonymous
  • anonymous
product rule?
anonymous
  • anonymous
The product rule applied to take derivative of the second term!
anonymous
  • anonymous
i'm confused
anonymous
  • anonymous
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)
anonymous
  • anonymous
what do you do next?
anonymous
  • anonymous
Yep, now put the 2 results together :) y = b cos t + t² sin t y' = - bsint + 2t sint+ t² cost
anonymous
  • anonymous
Now you can make it neat by simplifying - bsint + 2t sint together.
anonymous
  • anonymous
okaay thnks you

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