(\partial)/(\partial y)(2x^2+4y^2+1/2)

解答

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

8 y

求解步骤

\frac{\partial}{\partial y} (2 x^{2} + 4 y^{2} + \frac{1}{2})

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} (2 x^{2}) + \frac{\partial}{\partial y} (4 y^{2}) + \frac{\partial}{\partial y} (\frac{1}{2})

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

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

\frac{\partial}{\partial y} (\frac{1}{2}) = 0

= 0 + 8 y + 0

化简

= 8 y

y^{^{'}} = 1 + y

\int_{0}^{1} \frac{1}{x - 1} d x

y^{^{'}} = \frac{( x y + 3 x - y - 3 )}{( x y - 2 x + 4 y - 8 )}

in v erse l a pl a ce \frac{4}{( 3 s - 1 ) ^{5}}

y^{^{'}} + \frac{y}{x} = 2 y^{2}