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

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

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

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

v.s Group TitleBest ResponseYou've already chosen the best response.0
but i understand what you did
 2 years 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! :)
 2 years ago

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

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

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

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

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

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

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

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

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

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

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

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

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