What is (\partial)/(\partial x)(64-8x^2-y^2) ?

The answer to (\partial)/(\partial x)(64-8x^2-y^2) is -16x

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(\partial)/(\partial x)(64-8x^2-y^2)

\frac{\partial}{\partial x} (64 - 8 x^{2} - y^{2})

- 16 x

Solution steps

\frac{\partial}{\partial x} (64 - 8 x^{2} - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (64) - \frac{\partial}{\partial x} (8 x^{2}) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (64) = 0

\frac{\partial}{\partial x} (8 x^{2}) = 16 x

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

= 0 - 16 x - 0

Simplify

= - 16 x

x \to 1 + lim (\frac{1 - x}{x ( x - 1 )})

\frac{d}{d x} (cos^{2} (3 π x))

x \to - 9 - lim (\frac{x + 8}{x + 9})

\int \frac{( x ^{2} + 1 )}{x ( x - 1 ) ^{3}} d x

\int sec (x) \frac{1}{x} d x

What is (\partial)/(\partial x)(64-8x^2-y^2) ?
The answer to (\partial)/(\partial x)(64-8x^2-y^2) is -16x