What is (\partial)/(\partial y)(x^3+3x^2y-y^4) ?

The answer to (\partial)/(\partial y)(x^3+3x^2y-y^4) is 3x^2-4y^3

(\partial)/(\partial y)(x^3+3x^2y-y^4)

\frac{\partial}{\partial y} (x^{3} + 3 x^{2} y - y^{4})

3 x^{2} - 4 y^{3}

Solution steps

\frac{\partial}{\partial y} (x^{3} + 3 x^{2} y - y^{4})

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^{2} y) - \frac{\partial}{\partial y} (y^{4})

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

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

\frac{\partial}{\partial y} (y^{4}) = 4 y^{3}

= 0 + 3 x^{2} - 4 y^{3}

Simplify

= 3 x^{2} - 4 y^{3}

y^{^{''}} - 0.2 y^{^{'}} + 0.01 y = 0, y (0) = 15, y^{^{'}} (0) = b

x d y - (2 x + 1) e^{- y} d x = 0

\int \frac{1}{( 3 x - 1 )} d x

\frac{\partial}{\partial x} (2 cos (3 x))

\frac{d y}{d x} + y^{5} x + 4 y = 0

What is (\partial)/(\partial y)(x^3+3x^2y-y^4) ?
The answer to (\partial)/(\partial y)(x^3+3x^2y-y^4) is 3x^2-4y^3