(\partial)/(\partial x)(cos(x)e^{2xy}*2x)

解答

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

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

求解步骤

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

将

y

视为常数

将常数提出:

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

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

使用微分乘法定则:

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

f = cos (x), g = e^{2 x y} x

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

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

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

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

化简

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

\int \frac{x ^{5}}{x} d x

\int \frac{x ^{2}}{49 + x ^{6}} d x

s im pl i f y y - x^{2}

n = 1 \sum \infty \frac{5}{n} - \frac{5}{n + 1}

\int_{0}^{\frac{π}{2}} 6 cos^{5} (x) d x