What is (\partial)/(\partial x)(x^2+y^2-9) ?

The answer to (\partial)/(\partial x)(x^2+y^2-9) is 2x

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(\partial)/(\partial x)(x^2+y^2-9)

\frac{\partial}{\partial x} (x^{2} + y^{2} - 9)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} - 9)

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} (9)

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

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

\frac{\partial}{\partial x} (9) = 0

= 2 x + 0 - 0

Simplify

= 2 x

\int sec^{2} (2 x t a) n^{6} 2 x d x

\frac{d}{d x} (\frac{3}{x} + e^{2 x} \cdot 2)

f (x) = x^{2} - 3, (2, 1)

n = 0 \sum \infty \frac{n + 1}{3}

f (x) = 2 sin (x)

What is (\partial)/(\partial x)(x^2+y^2-9) ?
The answer to (\partial)/(\partial x)(x^2+y^2-9) is 2x