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

The answer to (\partial)/(\partial x)(x-x^2y-y+xy^2) is y^2-2xy+1

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

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

y^{2} - 2 x y + 1

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (x) = 1

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

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

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

= 1 - 2 y x - 0 + y^{2}

Simplify

= y^{2} - 2 x y + 1

x \to 0 lim (x^{7} sin (\frac{1}{x}))

(9 - \frac{x}{9})^{- 7}

x = - 2, x = 1, y = 7 x, y = x^{2} - 8

\frac{d y}{d x}, 6 x^{2} + 3 x y + 2 y^{2} + 17 y - 6 = 0

y^{^{''}} - 7 y^{^{'}} + 12 y = 0

What is (\partial)/(\partial x)(x-x^2y-y+xy^2) ?
The answer to (\partial)/(\partial x)(x-x^2y-y+xy^2) is y^2-2xy+1