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

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

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

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

3 x^{2} - 4 y

Solution steps

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

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

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

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

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

Simplify

= 3 x^{2} - 4 y

\frac{1}{s ( s ^{2} - 2 s + 2 )}

y^{^{'}} + y + \frac{1}{x} y = \frac{e ^{- x}}{x}

f (x) = x^{\frac{9}{2}}

3 y^{^{''}} - 4 y^{^{'}} - 15 y = 0

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