(\partial)/(\partial y)(3x^2-2y^3+6xy^2-8)

解答

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

- 6 y^{2} + 12 x y

求解步骤

\frac{\partial}{\partial y} (3 x^{2} - 2 y^{3} + 6 x y^{2} - 8)

将

x

视为常数

使用微分加减法定则:

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

= \frac{\partial}{\partial y} (3 x^{2}) - \frac{\partial}{\partial y} (2 y^{3}) + \frac{\partial}{\partial y} (6 x y^{2}) - \frac{\partial}{\partial y} (8)

\frac{\partial}{\partial y} (3 x^{2}) = 0

\frac{\partial}{\partial y} (2 y^{3}) = 6 y^{2}

\frac{\partial}{\partial y} (6 x y^{2}) = 12 x y

\frac{\partial}{\partial y} (8) = 0

= 0 - 6 y^{2} + 12 x y - 0

化简

= - 6 y^{2} + 12 x y

t an g e n t 2 cos (x) + cos^{2} (x)

x \to - 2 - lim (\frac{1}{x + 2})

f (x) = 9 ln (x)

\frac{\partial}{\partial x} (\frac{2 x y ^{2}}{1 + x ^{2} y ^{2}})

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0