What is (\partial)/(\partial x)(x^2+2y^2-x^2y-1) ?

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

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(\partial)/(\partial x)(x^2+2y^2-x^2y-1)

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} - x^{2} y - 1)

2 x - 2 x y

Solution steps

\frac{\partial}{\partial x} (x^{2} + 2 y^{2} - x^{2} y - 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2}) + \frac{\partial}{\partial x} (2 y^{2}) - \frac{\partial}{\partial x} (x^{2} y) - \frac{\partial}{\partial x} (1)

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

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

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

\frac{\partial}{\partial x} (1) = 0

= 2 x + 0 - 2 y x - 0

Simplify

= 2 x - 2 x y

y^{^{''}} + 81 y = cos (4 t)

\frac{d}{d t} (\frac{1}{3} π r (t) (t)^{2} h)

\frac{\partial}{\partial x} (\frac{7 x - 5 y}{y ^{2} + x ^{2}})

x \to - π lim ((x + π) csc (x))

8 t \frac{d y}{d t} + y = t^{7}

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