(\partial)/(\partial y)({f}(x,y)(x+y))

解答

\frac{\partial}{\partial y} (f (x, y) (x + y))

\frac{\partial}{\partial y} (f (x, y)) (x + y) + 1 \cdot f (x, y)

求解步骤

\frac{\partial}{\partial y} (f (x, y) (x + y))

将

x

视为常数

使用微分乘法定则:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = f (x, y), g = x + y

= \frac{\partial}{\partial y} (f (x, y)) (x + y) + \frac{\partial}{\partial y} (x + y) f (x, y)

\frac{\partial}{\partial y} (x + y) = 1

= \frac{\partial}{\partial y} (f (x, y)) (x + y) + 1 \cdot f (x, y)

d er i v a t i v e f (x) = x + 13

x y^{^{'}} - y = - 2 x^{2} - 2 x + 1

\frac{\partial}{\partial z} (x y^{2} e^{y})

\frac{d}{d x} y = (3 - e^{- x})^{2}

\int \frac{1}{4 x - 10} d x