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

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

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

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

2 x y + x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (x^{2}) = 0

= 2 x y + x - 0

Simplify

= 2 x y + x

\frac{d}{d x} (x x^{2} + y^{2} + z^{2})

\frac{d}{d x} (4 x sin (π x))

f (x) = \frac{x ^{2}}{x + 8}, at x = 2

\int 10 x ln (9 x) d x

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