Question

find y'' if y=8xsinx

I am getting
y''=8xsinx+16cosx
But that is not one of the choices
What am I doing wrong?

Answer 1

what are the choices first?

Answer 2

u have to find double derivative right? what is the first derivative u got?

Answer 3

and yeah this looks wrong.. there should be 1 more term i think

Answer 4

\[A. y''=8cosx-16xsinx\]
\[B. y''=16cosx-8xsinx\]
\[C. y''=-16cosx+8xsinx\]
\[D. y''=-8xsinx\]

and yes I have to find the double derivative

Answer 5

well I got
y=8xsinx
then y' should be
\[y'=8x \frac{ d }{ dx }sinx+sinx \frac{ d }{ dx }(8x)\]
\[y'=8xcosx+8sinx\]

Answer 6

yes this is right..

Answer 7

well this looks like it can have the product rule used doesn't it

let f be 8x
and let g be sin x

you do f' times g + g' times f
so
8 times sinx = 8sinx



as i typed...you came to the correct way lol :)

Answer 8

so then I tried taking the derivative of y'
\[y''=(8x \frac{ d }{ dx }cosx+cosx \frac{ d }{ dx }(8x))+(8\frac{ d }{ dx }sinx+sinx \frac{ d }{ dx}(8))\]
which should equal
\[y''=8xsinx+cosx(8)+8cosx\]

Answer 9

nop.. derivative of cosx is -sinx.. thats the mistake ..

Answer 10

there we go ...as @nubeer has said

Answer 11

ok so it is
y''=-8xsinx+16cosx
which is choice B
right?

Answer 12

yes..

Answer 13

yes it is

Answer 14

thanks guys

Answer 15

your welcome :)

Answer 16

anytime..

Answer 17

I am still trying to figure out a volume of a rectangle one

Answer 18

word problem
volume of a rectangular box with a square base remains constant at 1000cm^3 as the area of the base increases at a rate of 2cm^2/sec. Find the rate at which the height of the box is decreasing when each side of the base is 19cm long

Answer 19

so if A*h=1000 and the given rate of change is 2cm^2/sec
then do I
\[A*\frac{ d }{ dt }(h)+h*\frac{ d }{ dt }(A)=\frac{ d }{ dt }1000\]

Answer 20

hmm no clue on this one sorry

Answer 21

thanks anyway