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

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

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

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

6 x^{2} y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 6 x^{2} y - 0

Simplify

= 6 x^{2} y

\int_{1}^{7} 3 x^{2} + \frac{1}{12 x ^{2}} d x

\frac{d}{d x} (ln^{2} (1 + x))

y^{^{'}} = 2 y - 1

\frac{d}{d x} (x^{\frac{4}{3}} (2 - x)^{3})

\int \frac{49 - x ^{2}}{x ^{4}} d x

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