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

解答

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

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

求解步骤

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

将

x, z

视为常数

使用指数法则:

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

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

使用连锁律:

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

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

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

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

化简

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

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

n = 1 \sum \infty \frac{5}{1 0 ^{n}}

d er i v a t i v e y = 14 x

\int_{1}^{5} \frac{x ^{2} + 2}{6 x - x ^{2}} d x

a re a y^{2} - 2 x = 2, x - y = 3

\int \frac{2}{x - 1} d x