What is (\partial)/(\partial x)(x^2+y^2-y^2x-7) ?

The answer to (\partial)/(\partial x)(x^2+y^2-y^2x-7) is 2x-y^2

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(\partial)/(\partial x)(x^2+y^2-y^2x-7)

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

2 x - y^{2}

Solution steps

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

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

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

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

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

\frac{\partial}{\partial x} (7) = 0

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

Simplify

= 2 x - y^{2}

What is (\partial)/(\partial x)(x^2+y^2-y^2x-7) ?
The answer to (\partial)/(\partial x)(x^2+y^2-y^2x-7) is 2x-y^2