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

The answer to (\partial)/(\partial y)(4x^3y^2-2xy^3) is 8x^3y-6xy^2

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

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

8 x^{3} y - 6 x y^{2}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (4 x^{3} y^{2}) = 8 x^{3} y

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

= 8 x^{3} y - 6 x y^{2}

x^{3} + 7 e^{x}

\frac{x d y}{d x} = y^{2} - 4

\frac{d y}{d x} = (\frac{y}{x} + \frac{1}{4})^{2}

y = (f (x))^{2}, at x = 0

4500 (1 - \frac{1}{50} t)^{2}

What is (\partial)/(\partial y)(4x^3y^2-2xy^3) ?
The answer to (\partial)/(\partial y)(4x^3y^2-2xy^3) is 8x^3y-6xy^2