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

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

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

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

9 x^{2} + 8 x y

Solution steps

\frac{\partial}{\partial x} (3 x^{3} + 4 x^{2} y - 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} (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} (y^{3}) = 0

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

Simplify

= 9 x^{2} + 8 x y

x \to + 0 + lim (arctan (ln (x)))

\frac{3 s + 7}{s ^{2} - 2 s - 3}

\frac{d}{d x} (ln (x e^{x} + 7))

\int_{0}^{2} \frac{1}{( x - 1 ) ^{\frac{1}{3}}} d x

\frac{d}{d y} (\frac{y ^{3}}{21} + \frac{7}{4 y})

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