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

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

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

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

2 v

Solution steps

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

Treat

u

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 2 v - 0

Simplify

= 2 v

x y^{^{'}} + y = - 7 x y^{2}, y (1) = - 6

\int 14 - (20 - 3 16 - 2 x) d x

y^{- 1}

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

\frac{d y}{d x} = 12 - y

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