i can't see at all how come

(d/dx)(y'/sin^2y)=cosy/siny

i would think that

= [-2(y'^2)cosy/siny]+[y''/sin^2y]

it's frustrating me please give me some help

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- Mar 28th 2010, 10:55 AMpepsitotal differentiation
i can't see at all how come

(d/dx)(y'/sin^2y)=cosy/siny

i would think that

= [-2(y'^2)cosy/siny]+[y''/sin^2y]

it's frustrating me please give me some help - Mar 28th 2010, 01:18 PMskeeter
- Mar 28th 2010, 01:24 PMHallsofIvy
- Mar 29th 2010, 07:11 AMpepsi