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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

\frac{d}{d x} (x^{2} arctan (ln (x)))

\int \frac{ln ( 4 x )}{4 x ^{2}} d x

e^{x^{2} - 9}

2 x^{4} - x^{3} + 6 x^{2}

a (t) = t^{3} - 12 t

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