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

The answer to (\partial)/(\partial y)(1/(x^2+y^2+z^2)) is -(2y)/((x^2+y^2+z^2)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Treat

x, z

as constants

Apply exponent rule:

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

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

Apply the chain rule:

- \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

Simplify

- \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}}

What is (\partial)/(\partial y)(1/(x^2+y^2+z^2)) ?
The answer to (\partial)/(\partial y)(1/(x^2+y^2+z^2)) is -(2y)/((x^2+y^2+z^2)^2)