1)
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.
Can someone help me understand how to find the derivative of these two problems:
(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...:(
3 answers
I guess it's the format but I am little confused about what you did....
1)
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
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
1 answer