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

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

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

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

6 x^{2} y

Solution steps

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

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} (4)

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

\frac{\partial}{\partial y} (4) = 0

= 6 x^{2} y + 0

Simplify

= 6 x^{2} y

\frac{\partial}{\partial x} (- 8 (x + y + z)^{2})

\int \frac{arcsin ( x )}{( 1 - x ^{2} ) ^{\frac{1}{2}}} d x

x \to 0 lim (x^{15 x})

\int \frac{x ^{3}}{6 - x ^{4}} d x

\frac{6 u ^{2}}{( u ^{2} + u ) ^{3}}

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