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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (x^{4}) = 0

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

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

Simplify

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

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