What is (\partial)/(\partial y)(3x^2y^2+1) ?

The answer to (\partial)/(\partial y)(3x^2y^2+1) is 6x^2y

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(\partial)/(\partial y)(3x^2y^2+1)

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 1)

6 x^{2} y

Solution steps

\frac{\partial}{\partial y} (3 x^{2} y^{2} + 1)

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} (1)

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

\frac{\partial}{\partial y} (1) = 0

= 6 x^{2} y + 0

Simplify

= 6 x^{2} y

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

\frac{x}{( 1 + x ^{2} )} d x + \frac{y}{( 1 + y ^{2} )} d y = 0

f (x) = \frac{2 x}{1 + x ^{2}}

63 x^{2} y^{^{''}} + 14 x (x - 9) y^{^{'}} - 14 (x - 9) y = - 2 x^{3}

y = x^{2} - x

What is (\partial)/(\partial y)(3x^2y^2+1) ?
The answer to (\partial)/(\partial y)(3x^2y^2+1) is 6x^2y