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

The answer to (\partial)/(\partial x)(16-4x^2-y^2) is -8x

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

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

- 8 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 x^{2}) = 8 x

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

= 0 - 8 x - 0

Simplify

= - 8 x

y^{^{''}} + y^{^{'}} + 1 = 0

\frac{d}{d x} (\frac{2 x + 3}{x ^{2} - 9})

y = x^{4} + 9 e^{x}, (0, 5)

\int h^{- \frac{1}{2}} d h

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

What is (\partial)/(\partial x)(16-4x^2-y^2) ?
The answer to (\partial)/(\partial x)(16-4x^2-y^2) is -8x