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

解答

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

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

求解步骤

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

将

y

视为常数

将常数提出:

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

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

使用指数法则:

\frac{1}{a} = a^{- 1}

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

使用连锁律:

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

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

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

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

化简

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

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

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