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

The answer to (\partial)/(\partial x)(2x^3+2x^2y-3y^3) is 6x^2+4xy

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

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

6 x^{2} + 4 x y

Solution steps

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

\frac{\partial}{\partial x} (2 x^{3}) = 6 x^{2}

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

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

= 6 x^{2} + 4 y x - 0

Simplify

= 6 x^{2} + 4 x y

\int_{- 10}^{10} 100 - x^{2} d x

\frac{d y}{d x} = x^{2} x - 2

x \to 1 lim (\frac{- 5}{1 - x})

\int 3 (2 x + 5)^{3} d x

y = ln (\frac{5 x}{7 + x})

What is (\partial)/(\partial x)(2x^3+2x^2y-3y^3) ?
The answer to (\partial)/(\partial x)(2x^3+2x^2y-3y^3) is 6x^2+4xy