(\partial)/(\partial x)(1(x-y)*e^{1y+4x^2})

解答

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

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

求解步骤

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

化简

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

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

将

y

视为常数

使用微分乘法定则:

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

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

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

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

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

= e^{y + 4 x^{2}} \cdot 8 x (x - y) + 1 \cdot e^{y + 4 x^{2}}

化简

= 8 e^{y + 4 x^{2}} x (x - y) + e^{y + 4 x^{2}}

\frac{d ^{3}}{d x ^{3}} (6 x + 3)

\int_{0}^{s i n (x)} 8 t d t

\int - x - 3 d x

d er i v a t i v e h (x) = (x^{2} - 3)^{3}

\int_{- 8}^{2} \frac{1}{∣ 2 x ∣} d x