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

The answer to (\partial)/(\partial z)(ln((xy)/(y^2+z^2))) is -(2z)/(y^2+z^2)

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(\partial)/(\partial z)(ln((xy)/(y^2+z^2)))

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

- \frac{2 z}{y ^{2} + z ^{2}}

Solution steps

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

Treat

x, y

as constants

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{2 z}{y ^{2} + z ^{2}}

y^{2} = x, y = x - 2

\frac{d}{d x} (x + y + e^{x y})

y^{^{'}} - y = e^{3 t}

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

y^{^{'}} = (x - 4) e^{- 2 y}

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