What is (\partial)/(\partial y)(16-4x^2-y^2) ?

The answer to (\partial)/(\partial y)(16-4x^2-y^2) is -2y

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(\partial)/(\partial y)(16-4x^2-y^2)

\frac{\partial}{\partial y} (16 - 4 x^{2} - y^{2})

- 2 y

Solution steps

\frac{\partial}{\partial y} (16 - 4 x^{2} - y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (16) - \frac{\partial}{\partial y} (4 x^{2}) - \frac{\partial}{\partial y} (y^{2})

\frac{\partial}{\partial y} (16) = 0

\frac{\partial}{\partial y} (4 x^{2}) = 0

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

= 0 - 0 - 2 y

Simplify

= - 2 y

What is (\partial)/(\partial y)(16-4x^2-y^2) ?
The answer to (\partial)/(\partial y)(16-4x^2-y^2) is -2y