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

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

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

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

x^{2} - 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= x^{2} - 2 y

\frac{\partial}{\partial x} (\frac{2 s - 3 t}{3 s + 5 t})

\int \frac{1}{( x ^{2} + 9 ) ^{3}} d x

(x^{4} + 5 x^{2} - 5)^{4}

\frac{\partial}{\partial x} (2 + x ln (x y - 9))

u^{1}

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