What is (\partial)/(\partial x)(3x^3+4x^2y-3y^3) ?

The answer to (\partial)/(\partial x)(3x^3+4x^2y-3y^3) is 9x^2+8xy

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(\partial)/(\partial x)(3x^3+4x^2y-3y^3)

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 3 y^{3})

9 x^{2} + 8 x y

Solution steps

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 3 y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (3 x^{3}) + \frac{\partial}{\partial x} (4 x^{2} y) - \frac{\partial}{\partial x} (3 y^{3})

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

\frac{\partial}{\partial x} (4 x^{2} y) = 8 y x

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

= 9 x^{2} + 8 y x - 0

Simplify

= 9 x^{2} + 8 x y

x \to 0 lim (\frac{3 8 + x - 3 8}{x})

\int \frac{x + 5}{x ^{2} + 10 x + 26} d x

\frac{d y}{d t} = 3 t^{2} cos (t^{3})

x^{- x}

\frac{d}{d x} (5 x^{2} cos (x) cot (x))

What is (\partial)/(\partial x)(3x^3+4x^2y-3y^3) ?
The answer to (\partial)/(\partial x)(3x^3+4x^2y-3y^3) is 9x^2+8xy