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

解

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

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

解答ステップ

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

x, z

を定数として扱う

指数の規則を適用する:

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

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

連鎖法則を適用:

- \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

簡素化

- \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}}