Question

英语数学题求解~

已求出dy/dx=(x^-2xy^2-2x^3)/(2yx^2+2y^3+y)..求之后的详细步骤。。。谢谢

Answer 1

y' =(x-2xy²-2x³)/(2yx²+2y³+y)
(1) y' = 0, x-2xy²-2x³ = x(1 - 2y² - 2x²) = 0
(a) x = 0 (discard, prove yourself)
(b) 1 - 2y² - 2x² = 0, x² + y² = 1/2 (i)
(x² + y²)² = x² - y² becomes x² - y² = 1/4 (ii)
(i)+(ii): x² = 3/8, x = ±√6/4
(i)-(ii): y² = 1/8, y = ±√2/4
At the following 4 points, the tangent is horizontal:
(√6/4, √2/4), (√6/4, -√2/4), (-√6/4, √2/4), (-√6/4, -√2/4)

For the points,wher the tangent is vertical, y' doesn't exist, which means 2yx²+2y³+y = 0
2yx²+2y³+y = y(2x²+2y²+1)
Clearly, 2x²+2y²+1 > 0, which means y = 0
(x² + y²)² = x² - y² becomes x⁴ = x²
x²(x² - 1) = 0
x = 0 (discard)
x = ±1
The 2 points are: (1, 0), (-1,0)

Answer 2

Where' your mathematics problem?

追问

已求出dy/dx=(x^-2xy^2-2x^3)/(2yx^2+2y^3+y)..求之后的详细步骤。。。谢谢