What is (\partial)/(\partial x)(x^3-9xy+y^3-2) ?

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

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(\partial)/(\partial x)(x^3-9xy+y^3-2)

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

3 x^{2} - 9 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) - \frac{\partial}{\partial x} (9 x y) + \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (2)

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

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

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

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

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

Simplify

= 3 x^{2} - 9 y

\int_{1}^{e} \frac{sec ( 2 + ln ( x ) )}{x} d x

\int e^{2 x + l n (x)} d x

x \to 7 - lim (\frac{1}{x ^{2} ( x + 7 )})

\int \frac{2 x ^{3}}{2 x ^{2} - 4 x + 3} d x

\int x^{3} d x

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