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

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

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

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

6 y - 2 x y

Solution steps

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

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

\frac{\partial}{\partial y} (x^{3}) = 0

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

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

= 0 - 2 x y + 6 y

Simplify

= 6 y - 2 x y

y = 4 x^{3} - 2 x, (1, 2)

f (x) = x^{2} + 1, (- 4, 17)

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

\frac{z ^{\frac{8}{3}}}{8} - \frac{e ^{- 2 z}}{2}

\frac{d}{d x} (\frac{3 x + 4}{2 x - 5})

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