(\partial)/(\partial y)(xe^{y^2}+yln(x))

解答

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

x e^{y^{2}} \cdot 2 y + ln (x)

求解步骤

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

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

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

\frac{\partial}{\partial y} (y ln (x)) = ln (x)

= x e^{y^{2}} \cdot 2 y + ln (x)

\int \frac{6}{( x - 4 ) ( x + 2 )} d x

y^{^{''}} + 81 y = sec^{2} (9 x)

s l o p e (\frac{1}{4}, \frac{7}{8}), (4 \frac{1}{4}, 2 \frac{7}{8})

n = 0 \sum \infty \frac{1}{( 2 n + 1 ) ^{6}}

y d x + (1 + x) d y = 0