What is (\partial)/(\partial x)(12x^4-6x^2y+4y^3) ?

The answer to (\partial)/(\partial x)(12x^4-6x^2y+4y^3) is 48x^3-12xy

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(\partial)/(\partial x)(12x^4-6x^2y+4y^3)

\frac{\partial}{\partial x} (12 x^{4} - 6 x^{2} y + 4 y^{3})

48 x^{3} - 12 x y

Solution steps

\frac{\partial}{\partial x} (12 x^{4} - 6 x^{2} y + 4 y^{3})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (12 x^{4}) - \frac{\partial}{\partial x} (6 x^{2} y) + \frac{\partial}{\partial x} (4 y^{3})

\frac{\partial}{\partial x} (12 x^{4}) = 48 x^{3}

\frac{\partial}{\partial x} (6 x^{2} y) = 12 y x

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

= 48 x^{3} - 12 y x + 0

Simplify

= 48 x^{3} - 12 x y

\frac{d}{d x} (x^{\frac{3}{5}} (4 - x))

arccos (e^{2 x})

\int_{0}^{2} \frac{1}{x ^{1.9}} d x

f (x) = \frac{x + 3}{x + 5}

\frac{\partial}{\partial u} (3 u^{2})

What is (\partial)/(\partial x)(12x^4-6x^2y+4y^3) ?
The answer to (\partial)/(\partial x)(12x^4-6x^2y+4y^3) is 48x^3-12xy