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anonymous
 3 years ago
Differentiate with respect to t.
y = b cos t + t2 sin t
anonymous
 3 years ago
Differentiate with respect to t. y = b cos t + t2 sin t

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sin t (2b) + 2t cos t

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0> b ( cos t ) ' =  bsint Okie, so far?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y=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(2b)+2tcost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0but i understand what you did

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0datz 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
 3 years ago
Best ResponseYou've already chosen the best response.0Apply product rule here ( t²* sin t ) ' with : u = t² > u' = ... v = sint > v' = ....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@v.s Do you know product rule?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, now put the 2 results together :) y' = ....

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = b cos t + t² sin t >y' = ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y"= bcost+2t*sint+cost*t^2

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = b cos t + t² sin t > b ( cos t ) ' =  bsint > ( t² sin t )' = 2t sint+ t² cost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you just put them together

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0 bsint + 2t sint+ t² cost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What's my instruction above?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The product rule applied to take derivative of the second term!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0y = 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
 3 years ago
Best ResponseYou've already chosen the best response.0Yep, now put the 2 results together :) y = b cos t + t² sin t y' =  bsint + 2t sint+ t² cost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now you can make it neat by simplifying  bsint + 2t sint together.
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