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

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

Popular >

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

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

3 y^{2} - 3 x

Solution steps

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

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

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

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

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

= 0 - 3 x + 3 y^{2}

Simplify

= 3 y^{2} - 3 x

x^{2} sin^{3} (x) + x cos^{- 3} (x)

x \to \infty lim (x^{2} + 6)

f (x) = (4 x + 4)^{\frac{1}{3}}, at x = 1

\int 10 x - x^{2} d x

\int (x^{\frac{2}{3}} + 3) d x

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