What is (\partial)/(\partial x)(x^3-9xy+7y^3) ?

The answer to (\partial)/(\partial x)(x^3-9xy+7y^3) is 3x^2-9y

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(\partial)/(\partial x)(x^3-9xy+7y^3)

\frac{\partial}{\partial x} (x^{3} - 9 x y + 7 y^{3})

3 x^{2} - 9 y

Solution steps

\frac{\partial}{\partial x} (x^{3} - 9 x y + 7 y^{3})

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} (7 y^{3})

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

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

\frac{\partial}{\partial x} (7 y^{3}) = 0

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

Simplify

= 3 x^{2} - 9 y

\frac{d}{d x} (750 + 12 x - 0.8 x^{7})

cos (x^{4})

\frac{d}{d x} ((ln (2 x)) d x)

x \cdot e^{- x}

\int 35 x^{4} d x

What is (\partial)/(\partial x)(x^3-9xy+7y^3) ?
The answer to (\partial)/(\partial x)(x^3-9xy+7y^3) is 3x^2-9y