(\partial)/(\partial y)(2xyln(y))

解答

\frac{\partial}{\partial y} (2 x y ln (y))

2 x (ln (y) + 1)

求解步骤

\frac{\partial}{\partial y} (2 x y ln (y))

将

x

视为常数

将常数提出:

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

= 2 x \frac{\partial}{\partial y} (y ln (y))

使用微分乘法定则:

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

f = y, g = ln (y)

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

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

\frac{\partial}{\partial y} (ln (y)) = \frac{1}{y}

= 2 x (1 \cdot ln (y) + \frac{1}{y} y)

化简

2 x (1 \cdot ln (y) + \frac{1}{y} y) : 2 x (ln (y) + 1)

= 2 x (ln (y) + 1)

s im pl i f y \frac{5}{12} (3 + 5) x^{3}

\frac{\partial}{\partial v} (3 v cos (u))

d er i v a t i v e \frac{( 1 + x )}{x}

a re a y = \frac{1}{x ^{2} + 1}, x = 0, x = 6, y = 0

\frac{d}{d x} (- 6 x^{2} + 5)