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)

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(\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}}

\frac{d}{d x} (2 - tan (x))

\int \frac{lo g _{e} ( x )}{x} d x

\frac{\partial}{\partial x} (17 - x^{2} - y^{2})

\frac{d}{d x} (\frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}})

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

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)