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

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

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

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

- 2 x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 - 2 x - 0

Simplify

= - 2 x

x \to 1 lim (x - x)

\frac{\partial}{\partial x} (4 x^{- 1})

f (x) = x^{2} + 6 x, at a = 3

x \to 2 lim (3 x - 2)

\frac{x ^{2}}{8 + x}

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