What is (\partial)/(\partial x)(x^2+y^2-y^2x-8) ?

The answer to (\partial)/(\partial x)(x^2+y^2-y^2x-8) is 2x-y^2

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(\partial)/(\partial x)(x^2+y^2-y^2x-8)

\frac{\partial}{\partial x} (x^{2} + y^{2} - y^{2} x - 8)

2 x - y^{2}

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} - y^{2} x - 8)

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} (8)

\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} (8) = 0

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

Simplify

= 2 x - y^{2}

\int 2 x - \frac{2}{x} d x

\frac{\partial}{\partial y} (x arctan (\frac{y}{z}))

x \to + (- 3) lim (2 x + 5)

g (u) = 2 u + 3 u

x \to 0 lim (- 2 x + 1)

What is (\partial)/(\partial x)(x^2+y^2-y^2x-8) ?
The answer to (\partial)/(\partial x)(x^2+y^2-y^2x-8) is 2x-y^2