What is (\partial)/(\partial x)(81-9x^2-y^2) ?

The answer to (\partial)/(\partial x)(81-9x^2-y^2) is -18x

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(\partial)/(\partial x)(81-9x^2-y^2)

\frac{\partial}{\partial x} (81 - 9 x^{2} - y^{2})

- 18 x

Solution steps

\frac{\partial}{\partial x} (81 - 9 x^{2} - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (81) - \frac{\partial}{\partial x} (9 x^{2}) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (81) = 0

\frac{\partial}{\partial x} (9 x^{2}) = 18 x

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

= 0 - 18 x - 0

Simplify

= - 18 x

y^{2} = 5 x^{4} - x^{2}, (- 1, 2)

\int \frac{1}{y ^{2} ( y ^{2} + 1 )} d y

M^{2} (\frac{C}{2} - \frac{M}{3})

y^{^{'}} = 0.002 y - 200, y (0) = 80000

e^{4 x^{2}}

What is (\partial)/(\partial x)(81-9x^2-y^2) ?
The answer to (\partial)/(\partial x)(81-9x^2-y^2) is -18x