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標題:
If x^3 + y^3 = 1, find an expression for y".?
發問:
請問圖中計算正確嗎?有什麼問題? Thanks!! 更新: 謝謝大家的解答!!
最佳解答:
By EXPLICIT DIFFERENTIATION : - [ Your L.H.S. sol. ] . . .----------------------------------------- x3+ y3 = 1 y3 = 1-x3 y = (1-x3)^(1/3) - - - - - - - (#) (a) i) If y ' is expressed in terms of x : then y ' = [(1/3)(1-x3)^(-2/3)]*(-3x2) = -x2 (1-x3)^(-2/3) ii) If y ' is to be expressed in terms of x, y : then y ' = -x2 (1-x3)^(-2/3) = -x2 [(1-x3)^(1/3)]?2 = -x2 y?2 . . . . . . . (from #) (b) i) If y ' is expressed in terms of x : then y '' = -x2 d/dx[(1-x3)^(-2/3)] - (1-x3)^(-2/3) d/dx(x2) = -x2 (-2/3) * (1-x3)^(-5/3) - 2x * (1-x3)^(-2/3) = (2/3)x (1-x3)^(-5/3) [x - 3(1-x3)] = (2/3)x (1-x3)^(-5/3) * (-3+x+3x3) = (2x/3)(-3+x+3x3) * (1-x3)^(-5/3) ii) If y '' is to be expressed in terms of x, y : then y '' = (2x/3)(-3+x+3x3) * [(1-x3)^(1/3)]?? = (2x/3)(-3+x+3x3) * y?? . . . . . . . (from #) _________________________________________________________________________ By IMPLICIT DIFFERENTIATION : - [ Your R.H.S. sol. ] . . .----------------------------------------- i) x3+ y3 = 1 d/dx (x3 + y3) = d/dx (1) d/dx (x3) + d/dx (y3) = 0 3x2 + 3y2 (dy/dx) = 0 x2 + y2 y ' = 0 . . . . . . . . . . . [ ∵ dy/dx = y ' ] ∴ y ' = - x2/y2 但這方法是 Form 6 - U1 level,你可能未學！ ii) d/dx (y ') =d/dx(- x2/y2) . . . . . . . . . . . . . . . . . .
其他解答:
If x^3 + y^3 = 1, find an expression for y".?
發問:
請問圖中計算正確嗎?有什麼問題? Thanks!! 更新: 謝謝大家的解答!!
最佳解答:
By EXPLICIT DIFFERENTIATION : - [ Your L.H.S. sol. ] . . .----------------------------------------- x3+ y3 = 1 y3 = 1-x3 y = (1-x3)^(1/3) - - - - - - - (#) (a) i) If y ' is expressed in terms of x : then y ' = [(1/3)(1-x3)^(-2/3)]*(-3x2) = -x2 (1-x3)^(-2/3) ii) If y ' is to be expressed in terms of x, y : then y ' = -x2 (1-x3)^(-2/3) = -x2 [(1-x3)^(1/3)]?2 = -x2 y?2 . . . . . . . (from #) (b) i) If y ' is expressed in terms of x : then y '' = -x2 d/dx[(1-x3)^(-2/3)] - (1-x3)^(-2/3) d/dx(x2) = -x2 (-2/3) * (1-x3)^(-5/3) - 2x * (1-x3)^(-2/3) = (2/3)x (1-x3)^(-5/3) [x - 3(1-x3)] = (2/3)x (1-x3)^(-5/3) * (-3+x+3x3) = (2x/3)(-3+x+3x3) * (1-x3)^(-5/3) ii) If y '' is to be expressed in terms of x, y : then y '' = (2x/3)(-3+x+3x3) * [(1-x3)^(1/3)]?? = (2x/3)(-3+x+3x3) * y?? . . . . . . . (from #) _________________________________________________________________________ By IMPLICIT DIFFERENTIATION : - [ Your R.H.S. sol. ] . . .----------------------------------------- i) x3+ y3 = 1 d/dx (x3 + y3) = d/dx (1) d/dx (x3) + d/dx (y3) = 0 3x2 + 3y2 (dy/dx) = 0 x2 + y2 y ' = 0 . . . . . . . . . . . [ ∵ dy/dx = y ' ] ∴ y ' = - x2/y2 但這方法是 Form 6 - U1 level,你可能未學！ ii) d/dx (y ') =d/dx(- x2/y2) . . . . . . . . . . . . . . . . . .
其他解答:
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圖中兩個也是錯誤的，左方第一次微分開始錯誤的，而右方的d/dy是不能像平時的做分數一樣 x^3+y^3=1 y^3=1-x^3 (3y^2)(dy/dx)=-3x^2 (y^2)(dy/dx)=-x^2 (2y)(d^2 y/dx^2)=-2x dy^2 /dx^2=-x/y y''=-x/y
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