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Shriya Group TitleBest ResponseYou've already chosen the best response.1
sin t (2b) + 2t cos t
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
( cost)' ? (sint)' ?
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
> b ( cos t ) ' =  bsint Okie, so far?
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
y' = bsin+2tcos
 one year ago

Shriya Group TitleBest ResponseYou've already chosen the best response.1
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(2b)+2tcost
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
i got it wrong
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
but i understand what you did
 one year ago

Shriya Group TitleBest ResponseYou've already chosen the best response.1
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! :)
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
yeaaa thank you :)
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Apply product rule here ( t²* sin t ) ' with : u = t² > u' = ... v = sint > v' = ....
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
@v.s Do you know product rule?
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
2t*sint + cost*t^2
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Yep, now put the 2 results together :) y' = ....
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
y = b cos t + t² sin t >y' = ...
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
y"= bcost+2t*sint+cost*t^2
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
y = b cos t + t² sin t > b ( cos t ) ' =  bsint > ( t² sin t )' = 2t sint+ t² cost
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
you just put them together
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
 bsint + 2t sint+ t² cost
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
What's my instruction above?
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
The product rule applied to take derivative of the second term!
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
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)
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
what do you do next?
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Yep, now put the 2 results together :) y = b cos t + t² sin t y' =  bsint + 2t sint+ t² cost
 one year ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
Now you can make it neat by simplifying  bsint + 2t sint together.
 one year ago

v.s Group TitleBest ResponseYou've already chosen the best response.0
okaay thnks you
 one year ago
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