derivative of x^4+y^4=4,x^3+4y^3y^'=0

解

\frac{d}{d x} (x^{4} + y^{4}) = 4, x^{3} + 4 y^{3} y^{^{'}} = 0

y = 4 x + c_{1} - x^{4}, y = - 4 x + c_{1} - x^{4}, y = i 4 4 x + c_{1} - x^{4}, y = - i 4 4 x + c_{1} - x^{4}

解答ステップ

(x^{4} + y^{4})^{'} = 4

f^{^{'}} (x) = g (x)

であれば

f (x) = \int g (x) d x

x^{4} + y^{4} = \int 4 d x

\int 4 d x = 4 x + c_{1}

x^{4} + y^{4} = 4 x + c_{1}

分離する

y : y = 4 x + c_{1} - x^{4}, y = - 4 x + c_{1} - x^{4}, y = i 4 4 x + c_{1} - x^{4}, y = - i 4 4 x + c_{1} - x^{4}

y = 4 x + c_{1} - x^{4}, y = - 4 x + c_{1} - x^{4}, y = i 4 4 x + c_{1} - x^{4}, y = - i 4 4 x + c_{1} - x^{4}