Question

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

Answer 1

sin t (2-b) + 2t cos t

Answer 2

- sint

Answer 3

( cost)' ? (sint)' ?

Answer 4

-> b ( cos t ) ' = - bsint
Okie, so far?

Answer 5

okayy

Answer 6

y' = -bsin+2tcos

Answer 7

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

Answer 8

i got it wrong

Answer 9

but i understand what you did

Answer 10

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! :)

Answer 11

yeaaa thank you :)

Answer 12

Apply product rule here ( t²* sin t ) ' with :
u = t² --> u' = ...
v = sint --> v' = ....

Answer 13

@v.s Do you know product rule?

Answer 14

yes

Answer 15

2t*sint + cost*t^2

Answer 16

Yep, now put the 2 results together :)
y' = ....

Answer 17

y = b cos t + t² sin t
->y' = ...

Answer 18

y"= -bcost+2t*sint+cost*t^2

Answer 19

y = b cos t + t² sin t
-> b ( cos t ) ' = - bsint
-> ( t² sin t )' = 2t sint+ t² cost

Answer 20

you just put them together

Answer 21

- bsint + 2t sint+ t² cost

Answer 22

What's my instruction above?

Answer 23

product rule?

Answer 24

The product rule applied to take derivative of the second term!

Answer 25

i'm confused

Answer 26

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)

Answer 27

what do you do next?

Answer 28

Yep, now put the 2 results together :)
y = b cos t + t² sin t
y' = - bsint + 2t sint+ t² cost

Answer 29

Now you can make it neat by simplifying - bsint + 2t sint together.

Answer 30

okaay
thnks you