(\partial)/(\partial x)(4(x-y)e^{4y+2x^2})

解答

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

4 (e^{4 y + 2 x^{2}} + 4 e^{4 y + 2 x^{2}} x (x - y))

求解步骤

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

将

y

视为常数

将常数提出:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 \frac{\partial}{\partial x} ((x - y) e^{4 y + 2 x^{2}})

使用微分乘法定则:

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

f = x - y, g = e^{4 y + 2 x^{2}}

= 4 (\frac{\partial}{\partial x} (x - y) e^{4 y + 2 x^{2}} + \frac{\partial}{\partial x} (e^{4 y + 2 x^{2}}) (x - y))

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

\frac{\partial}{\partial x} (e^{4 y + 2 x^{2}}) = e^{4 y + 2 x^{2}} \cdot 4 x

= 4 (1 \cdot e^{4 y + 2 x^{2}} + e^{4 y + 2 x^{2}} \cdot 4 x (x - y))

化简

= 4 (e^{4 y + 2 x^{2}} + 4 e^{4 y + 2 x^{2}} x (x - y))

d er i v a t i v e f (x) = x^{3} - 1

\frac{\partial}{\partial x} (4 x^{3} - 5 x^{2} + 8 x)

y^{^{'}} = \frac{2 x + y}{2 x + y + 3}

\int sin^{2} (5 x) d x

d er i v a t i v e 4 e^{x} + \frac{8}{3 x}