(\partial)/(\partial z)(-9zcos(3x+yz))

解答

\frac{\partial}{\partial z} (- 9 z cos (3 x + yz))

- 9 (cos (3 x + zy) - zy sin (3 x + zy))

求解步骤

\frac{\partial}{\partial z} (- 9 z cos (3 x + yz))

将

x, y

视为常数

将常数提出:

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

= - 9 \frac{\partial}{\partial z} (z cos (3 x + yz))

使用微分乘法定则:

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

f = z, g = cos (3 x + yz)

= - 9 (\frac{\partial}{\partial z} (z) cos (3 x + yz) + \frac{\partial}{\partial z} (cos (3 x + yz)) z)

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

\frac{\partial}{\partial z} (cos (3 x + yz)) = - sin (3 x + yz) y

= - 9 (1 \cdot cos (3 x + yz) + (- sin (3 x + yz) y) z)

化简

= - 9 (cos (3 x + zy) - zy sin (3 x + zy))

\frac{d}{d x} (1 + x^{2} + y^{2})

\frac{d y}{d x} = 2 y (1 - y)

d er i v a t i v e \frac{x ^{2} + 5 x + 6}{x ^{2} - 4}

d er i v a t i v e \frac{2}{( 6 x ^{2} + 5 x + 1 ) ^{2}}

\int \frac{1}{aa uu} d u