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

\int sec^{2} (5 t) d t

g (x) = x^{2} (1 - 5 x)

f (x) = \frac{2 x ^{2} - 1}{x ^{3} - x}

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

x \to - 8 + lim (\frac{x}{x + 8})

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