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

The answer to (\partial)/(\partial z)((x^2-y^2)/(z^2)) is -(2(x^2-y^2))/(z^3)

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

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

- \frac{2 ( x ^{2} - y ^{2} )}{z ^{3}}

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= (x^{2} - y^{2}) (- 2 z^{- 2 - 1})

Simplify

(x^{2} - y^{2}) (- 2 z^{- 2 - 1}) : - \frac{2 ( x ^{2} - y ^{2} )}{z ^{3}}

= - \frac{2 ( x ^{2} - y ^{2} )}{z ^{3}}

\frac{d}{d x} (\frac{( 6 - sec ( x ) )}{tan ( x )})

n = 1 \sum \infty (\frac{3}{2})^{n}

\int \frac{6 x + 4}{3 x ^{2} + 4 x + 5} d x

x^{3} - 1, x = - 1, x = 2

\frac{d}{d x} (\frac{sec ^{2} ( x )}{2})

What is (\partial)/(\partial z)((x^2-y^2)/(z^2)) ?
The answer to (\partial)/(\partial z)((x^2-y^2)/(z^2)) is -(2(x^2-y^2))/(z^3)