What is (\partial)/(\partial u)(u^2+v^2-4) ?

The answer to (\partial)/(\partial u)(u^2+v^2-4) is 2u

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(\partial)/(\partial u)(u^2+v^2-4)

\frac{\partial}{\partial u} (u^{2} + v^{2} - 4)

2 u

Solution steps

\frac{\partial}{\partial u} (u^{2} + v^{2} - 4)

Treat

v

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial u} (u^{2}) + \frac{\partial}{\partial u} (v^{2}) - \frac{\partial}{\partial u} (4)

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

\frac{\partial}{\partial u} (v^{2}) = 0

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

= 2 u + 0 - 0

Simplify

= 2 u

\int x \cdot (x^{2} - 3) d x

y = x sin (\frac{1}{x})

3 x^{2} (x^{3} + 1)^{7}

x \to 0 lim (\frac{sin ^{2} ( 4 x )}{x})

x \to 0 + lim (- \frac{1}{sin ( 2 x )})

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