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

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

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

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

\frac{2 y}{x ^{3}} + 2 x y + 3 x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= \frac{2 y}{x ^{3}} + 2 x y + 3 x

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