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

The answer to (\partial)/(\partial y)(x^3-y^2+2xy) is 2x-2y

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(\partial)/(\partial y)(x^3-y^2+2xy)

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

2 x - 2 y

Solution steps

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

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

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

\frac{d y}{d x} = 8 y e^{x + 2}

\int \frac{1}{t ^{3} + t ^{2} - 2 t} d t

\int \frac{32 x ^{2}}{( x - 15 ) ( x + 5 ) ^{2}} d x

\int 8^{- x} d x

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