What is (\partial)/(\partial u)(2ucos(v)) ?

The answer to (\partial)/(\partial u)(2ucos(v)) is 2cos(v)

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(\partial)/(\partial u)(2ucos(v))

\frac{\partial}{\partial u} (2 u cos (v))

2 cos (v)

Solution steps

\frac{\partial}{\partial u} (2 u cos (v))

Treat

v

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 cos (v) \frac{\partial}{\partial u} (u)

Apply the common derivative:

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

= 2 cos (v) \cdot 1

Simplify

= 2 cos (v)

y^{^{'}} + y - 4 cos (x) = 0

\frac{s}{( s + 2 ) ( s + 3 )}

\frac{d}{d x} ((\frac{3 x - 1}{x ^{2} + 3})^{2})

x \to 0 lim (\frac{1 - cos ( x )}{5 x ^{2}})

y^{^{'}} - y = 10 t e^{2 t}

What is (\partial)/(\partial u)(2ucos(v)) ?
The answer to (\partial)/(\partial u)(2ucos(v)) is 2cos(v)