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

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

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

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

- 2 z

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 0 - 2 z

Simplify

= - 2 z

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