What is (\partial)/(\partial z)(2x^2+3y^2-z^2) ?

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

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(\partial)/(\partial z)(2x^2+3y^2-z^2)

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

- 2 z

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial z} (3 y^{2}) = 0

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

= 0 + 0 - 2 z

Simplify

= - 2 z

\frac{1}{( 2 x + 1 ) ^{2}}

\int \frac{1}{x ^{2} 4 x ^{2} - 25} d x

t^{2} (5 t + 2)^{3}

\int e^{4 x - 7} d x

f (x) = cot^{2} (x)

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