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

解答

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

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

求解步骤

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

将

y

视为常数

使用指数法则:

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

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

使用连锁律:

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

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

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

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

化简

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

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