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

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

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

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

2 y

Solution steps

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

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 2 y - 0

Simplify

= 2 y

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

\int (3 x + 5) cos (\frac{x}{4}) d x

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

e^{\frac{x}{y}}

3 (e^{x} cos (x) - e^{x} sin (x))

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