What is (\partial)/(\partial x)(9xy-x^3-y^3-6) ?

The answer to (\partial)/(\partial x)(9xy-x^3-y^3-6) is 9y-3x^2

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(\partial)/(\partial x)(9xy-x^3-y^3-6)

\frac{\partial}{\partial x} (9 x y - x^{3} - y^{3} - 6)

9 y - 3 x^{2}

Solution steps

\frac{\partial}{\partial x} (9 x y - x^{3} - y^{3} - 6)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (9 x y) - \frac{\partial}{\partial x} (x^{3}) - \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (6)

\frac{\partial}{\partial x} (9 x y) = 9 y

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

\frac{\partial}{\partial x} (y^{3}) = 0

\frac{\partial}{\partial x} (6) = 0

= 9 y - 3 x^{2} - 0 - 0

Simplify

= 9 y - 3 x^{2}

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10

\frac{\partial}{\partial y} (ln (3 x + 5 y))

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What is (\partial)/(\partial x)(9xy-x^3-y^3-6) ?
The answer to (\partial)/(\partial x)(9xy-x^3-y^3-6) is 9y-3x^2