検索キーワード「(x-y)2 formula」に一致する投稿を関連性の高い順に表示しています。 日付順 すべての投稿を表示
検索キーワード「(x-y)2 formula」に一致する投稿を関連性の高い順に表示しています。 日付順 すべての投稿を表示

[ベスト] x^2 y^2 2xy 735393-X^(2)-y^(2)+2xy(dy)/(dx)=0

X 2 y 2 = (x – y) 2 2xy Algebraic identities The algebraic equations which are valid for all values of variables in them are called algebraic identities For x^2y^2=2xy, we get (by differentiating implicitly), dy/dx =1 That's the same as the derivative of a linear function with slope, 1 Hmmmmm Let's see If we have x^2y^2=2xy The we must also have x^22xy y^2=0 Factoring gets us (xy)^2 = 0 And the only way for that to happen is to have xy=0 So y=x and dy/dx =1 1 Answer Sorted by 1 Alpha identifies it as Legendre's equation and gives the solution y ( x) = c 1 x c 2 ( − x ( log ( 1 − x) / 2 − log ( x 1)) − 1) It offers step by step if you have the right account Share

How Do You Differentiate X 2 2xy Y 2 X 74 Socratic

How Do You Differentiate X 2 2xy Y 2 X 74 Socratic

X^(2)-y^(2)+2xy(dy)/(dx)=0

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