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

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

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

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

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

Solution steps

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

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

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

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

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

Simplify

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

\frac{d ^{2}}{d x ^{2}} (e^{6 x})

\int \frac{x ^{2}}{e ^{x^{3}}} d x

\frac{d}{d x} (ln (16))

x \to 6 lim (\frac{7 x - 4}{2 x - 8})

\int_{0}^{2} x e^{x} d x

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