(\partial)/(\partial x)(7y-y^3-4x^2y)

解答

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

- 8 y x

求解步骤

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

将

y

视为常数

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (7 y) - \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (4 x^{2} y)

\frac{\partial}{\partial x} (7 y) = 0

\frac{\partial}{\partial x} (y^{3}) = 0

\frac{\partial}{\partial x} (4 x^{2} y) = 8 y x

= 0 - 0 - 8 y x

化简

= - 8 y x

y^{5} y^{^{'}} = - 2 cos (π x)

\int ln (3 x) d x

l a pl a ce t r an s f or m δ (t + 1)

\int \frac{1}{x 81 x ^{2} + 1} d x

d er i v a t i v e \frac{1}{x ^{5}}