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

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

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

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

6 y^{3} x - 6 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 6 y^{3} x - 6 x

\int 2 u d u

\int 2 x - 4 d x

y = \frac{81}{x ^{4}}

f (x) = x^{4} - 4 x^{2} + 7, at x = 1

\frac{d}{d x} (\frac{e ^{8 x}}{3 x ^{4}})

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