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January 28, 2015

January 28, 2015

Posted by **please help!** on Sunday, June 24, 2012 at 7:54pm.

(t^2-1/t^2+2)^3=y and y=sin^32x?

I think with the first one, I could just foil it maybe and cancel but I don't know what to do with the ^3...and then the second one I haven't gotten the hang of how to eliminate those kinds of problems yet...:(

- calculus -
**bobpursley**, Sunday, June 24, 2012 at 8:04pm1)

y'=d/dt (uv^-1) where u= t^2-1 du=2dt

and v=(t^2+2)^3 dv= 3(t^2+2)^2 * (2t)=6t(t^2+2)^2

y'= v^-1 du -uv^-2 dv

y'=du/v -udv/v^2

then put in u,v du, dv and you have it.

- calculus -
**please help!**, Sunday, June 24, 2012 at 9:43pmI guess it's the format but I am little confused about what you did....

- calculus -
**Steve**, Monday, June 25, 2012 at 10:40am1)

use the quotient rule:

y = u/v

where u = t^2-1 and v = t^2+2

y' = (u'v - uv')/v^2

= [(2t)(t^2+2) - (t^2-1)(2t)]/(t^2+2)^2

= (2t^3+4t-2t^3+2t)/(t^2+2)^2

= 6t/(t^2+2)^2

2) use the chain rule:

y = u^3 where u = sin2x

y' = 3u^2 u'

= 3sin^2(2x) 2cos2x

= 6sin^2(2x) cos(2x)

or

3sin4x sin2x

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