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

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

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

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

6 x y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 6 y^{2} x - 0

Simplify

= 6 x y^{2}

(2 - x - 3 x^{3}) (7 + x^{5})

(s^{\frac{- 1}{2}} + 7 s) (1 - s^{- 1})

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

x \to - 3 lim (19 + x)

x^{2} e^{x} - 2 x e^{x} + 2 e^{x}

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