(\partial)/(\partial y)(e^{2y}-ycos(xy))

解答

\frac{\partial}{\partial y} (e^{2 y} - y cos (x y))

2 e^{2 y} - cos (y x) + y x sin (y x)

求解步骤

\frac{\partial}{\partial y} (e^{2 y} - y cos (x y))

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} (e^{2 y}) - \frac{\partial}{\partial y} (y cos (x y))

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

\frac{\partial}{\partial y} (y cos (x y)) = cos (x y) - x y sin (x y)

= e^{2 y} \cdot 2 - (cos (x y) - x y sin (x y))

化简

e^{2 y} \cdot 2 - (cos (x y) - x y sin (x y)) : 2 e^{2 y} - cos (y x) + y x sin (y x)

= 2 e^{2 y} - cos (y x) + y x sin (y x)

\int 7 e^{- 0.1 x} d x

\int sin^{2} (x) d x

\frac{d}{d x} (7 x^{2} cos (x) cot (x))

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

\int 3 x^{2} ln (x^{2} + 5) d x