How to find (implicit differetiation) dy/dx in xy=(x-y)^2?
回答 (4)
2017-09-12 4:57 pm
xy = (x - y)^2
xy = x^2 - 2xy + y^2
y + xy' = 2x - (2y + 2xy') + 2yy'
y + xy' = 2x - 2y - 2xy' + 2yy'
3xy' - 2yy' = 2x - 3y
y'(3x - 2y) = 2x - 3y
y' = (2x - 3y)/(3x - 2y)
xy = x^2 - 2xy + y^2
y + xy' = 2x - (2y + 2xy') + 2yy'
y + xy' = 2x - 2y - 2xy' + 2yy'
3xy' - 2yy' = 2x - 3y
y'(3x - 2y) = 2x - 3y
y' = (2x - 3y)/(3x - 2y)
2017-09-12 4:53 pm
xy = (x - y)²
xy = x² - 2xy + y²
x² - 3xy + y² = 0
(d/dx)(x² - 3xy + y²) = 0
2x - 3x(dy/dx) - 3y + 2y(dy/dx)
3x(dy/dx) - 2y(dy/dx) = 2x - 3y
(3x - 2y)(dy/dx) = 2x - 3y
dy/dx = (2x - 3y)/(3x - 2y)
xy = x² - 2xy + y²
x² - 3xy + y² = 0
(d/dx)(x² - 3xy + y²) = 0
2x - 3x(dy/dx) - 3y + 2y(dy/dx)
3x(dy/dx) - 2y(dy/dx) = 2x - 3y
(3x - 2y)(dy/dx) = 2x - 3y
dy/dx = (2x - 3y)/(3x - 2y)
2017-09-12 9:56 pm
x y = (x-y)^2
differentiate both sides with respect to x
y + x dy/dx = 2(x-y) (1 - dy/dx)
y + x dy/dx = 2 (x- x dy/dx -y + y dy/dx)
y + x dy/dx = 2x - 2x dy/dx - 2y + 2y dy/dx
x dy/dx + 2x dy/dx - 2y dy/dx = 2x - 2y -y
dy/dx (x +2x-2y) = 2x-3y
dy/dx (3x-2y) = 2x-3y
dy/dx = (2x-3y) /(3x-2y)
differentiate both sides with respect to x
y + x dy/dx = 2(x-y) (1 - dy/dx)
y + x dy/dx = 2 (x- x dy/dx -y + y dy/dx)
y + x dy/dx = 2x - 2x dy/dx - 2y + 2y dy/dx
x dy/dx + 2x dy/dx - 2y dy/dx = 2x - 2y -y
dy/dx (x +2x-2y) = 2x-3y
dy/dx (3x-2y) = 2x-3y
dy/dx = (2x-3y) /(3x-2y)
2017-09-12 4:49 pm
xy = (x - y)^2
xy = x^2 - 2xy + y^2
x^2 - 3xy + y^2 = 0
Now differentiate using the Product Rule on the middle term:
2x - 3x(dy/dx) - 3y + 2y(dy/dx) = 0
-3x(dy/dx) + 2y(dy/dx) = 3y - 2x
(dy/dx)(-3x + 2y) = 3y - 2x
dy/dx = (3y - 2x) / (2y - 3x)
xy = x^2 - 2xy + y^2
x^2 - 3xy + y^2 = 0
Now differentiate using the Product Rule on the middle term:
2x - 3x(dy/dx) - 3y + 2y(dy/dx) = 0
-3x(dy/dx) + 2y(dy/dx) = 3y - 2x
(dy/dx)(-3x + 2y) = 3y - 2x
dy/dx = (3y - 2x) / (2y - 3x)
收錄日期: 2021-04-18 17:51:49
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