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

\int 4 y sin (x y) d x

\frac{d}{d x} (A (- 2 e^{- 2 x} x + e^{- 2 x}))

\int_{0}^{1} 2 + 5 x^{2} d x

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

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

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