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

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

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

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

2 x - y

Solution steps

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

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

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

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

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

= 2 x - y - 0

Simplify

= 2 x - y

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\int_{\frac{1}{7}}^{5} 16 x ln (7 x) d x

\frac{d}{d x} ((x)^{\frac{1}{2}})

\int_{- 2}^{2} x^{3} cos (\frac{x}{2}) + \frac{1}{2} d x

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