What is (\partial)/(\partial x)(x^2+y^2-4) ?

The answer to (\partial)/(\partial x)(x^2+y^2-4) is 2x

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(\partial)/(\partial x)(x^2+y^2-4)

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4)

2 x

Solution steps

\frac{\partial}{\partial x} (x^{2} + y^{2} - 4)

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} (4)

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

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

\frac{\partial}{\partial x} (4) = 0

= 2 x + 0 - 0

Simplify

= 2 x

\int \frac{x ^{2}}{361 - x ^{2}} d x

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 9} e^{3 t}

\int \frac{4 x + 2}{x ^{2} + x + 5} d x

5 x - x^{2} (1.4)

\frac{\partial}{\partial x} (x^{2} - 4 y^{2})

What is (\partial)/(\partial x)(x^2+y^2-4) ?
The answer to (\partial)/(\partial x)(x^2+y^2-4) is 2x