(\partial)/(\partial x)(3zcos(4xy)+x-7y)

解答

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

- 12 zy sin (4 x y) + 1

求解步骤

\frac{\partial}{\partial x} (3 z cos (4 x y) + x - 7 y)

将

z, y

视为常数

使用微分加减法定则:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (3 z cos (4 x y)) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (7 y)

\frac{\partial}{\partial x} (3 z cos (4 x y)) = - 12 zy sin (4 y x)

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

\frac{\partial}{\partial x} (7 y) = 0

= - 12 zy sin (4 y x) + 1 - 0

化简

= - 12 zy sin (4 x y) + 1

d er i v a t i v e f (x) = \frac{1}{6 x ^{3}}

t an g e n t f (x) = \frac{2}{x - 6}, at x = 8

\int_{2}^{5} \frac{x}{x ^{2} + 4 x + 20} d x

d er i v a t i v e (x^{2} - 16 x + 64) (x^{2} + 64 x + 4)

\int \frac{sec ( \frac{1}{x} )}{x ^{2}} d x