(\partial)/(\partial x)((2xy^2)/(1+x^2y^2))

解答

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

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

求解步骤

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

将

y

视为常数

将常数提出:

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

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

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 2 y^{2} \frac{\frac{\partial}{\partial x} ( x ) ( 1 + x ^{2} y ^{2} ) - \frac{\partial}{\partial x} ( 1 + x ^{2} y ^{2} ) x}{( 1 + x ^{2} y ^{2} ) ^{2}}

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

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

= 2 y^{2} \frac{1 \cdot ( 1 + x ^{2} y ^{2} ) - 2 y ^{2} xx}{( 1 + x ^{2} y ^{2} ) ^{2}}

化简

2 y^{2} \frac{1 \cdot ( 1 + x ^{2} y ^{2} ) - 2 y ^{2} xx}{( 1 + x ^{2} y ^{2} ) ^{2}} : \frac{2 y ^{2} ( 1 - x ^{2} y ^{2} )}{( 1 + x ^{2} y ^{2} ) ^{2}}

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

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = 0

x^{^{''}} + 9 x = 27, x (0) = 4, x^{^{'}} (0) = 6

n = 1 \sum \infty \frac{2 n}{5 n - 4}

t an g e n t y = x^{3} - 3 x^{2} + 25

\frac{d}{d x} (arctan (sec (x) + tan (x)))