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

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

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

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

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

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (y^{2}) = 2 y

= \frac{1}{z ^{2}} (0 - 2 y)

Simplify

\frac{1}{z ^{2}} (0 - 2 y) : - \frac{2 y}{z ^{2}}

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

\frac{x}{x + \frac{d}{x}}

f (x) = (2 x^{5} - 3)^{8}

x \to + (- 2.4) lim (x)

y^{^{'}} - t y^{3} = 0

n = 7 \sum \infty \frac{1}{n}

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