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

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

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

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

2 x + 2 y

Solution steps

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

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

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

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

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

= 0 + 2 y + 2 x

Simplify

= 2 x + 2 y

\frac{( 6 s + 3 )}{s ^{4} + 5 s ^{2} + 4}

- sin (x) - \frac{1}{2}

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\int \frac{1}{8 - 4 sin ( x ) + 7 cos ( x )} d x

\int - x ln (x) d x

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