What is (\partial)/(\partial u)(uv-1) ?

The answer to (\partial)/(\partial u)(uv-1) is v

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(\partial)/(\partial u)(uv-1)

\frac{\partial}{\partial u} (uv - 1)

v

Solution steps

\frac{\partial}{\partial u} (uv - 1)

Treat

v

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial u} (uv) - \frac{\partial}{\partial u} (1)

\frac{\partial}{\partial u} (uv) = v

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

= v - 0

Simplify

= v

\frac{2 s}{( s - 1 ) ( s ^{2} + 25 )}

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

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

f (x) = (5 x^{2} - 7 x) e^{x}

f (x) = ln^{3} (x)

What is (\partial)/(\partial u)(uv-1) ?
The answer to (\partial)/(\partial u)(uv-1) is v