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

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

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(\partial)/(\partial y)(x^2+2xy^2+2y^2)

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

4 x y + 4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{2}) = 0

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

\frac{\partial}{\partial y} (2 y^{2}) = 4 y

= 0 + 4 x y + 4 y

Simplify

= 4 x y + 4 y

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

y = 2 - ∣ x ∣, y = x^{2}

\frac{1}{x} + 6

\frac{1}{s ^{2} - π}

\int \frac{15}{x ^{2} - 1} d x

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