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

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

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

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

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

Solution steps

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

Treat

y, x

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d ^{6}}{d x ^{6}} (x^{- 7})

\int (3 x + cos (x)) d x

\int \frac{1}{x ^{2} - 6 x + 25} d x

\frac{\partial}{\partial x} (- 6 x + 18)

\frac{d}{d x} (x^{2} + 16)

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