What is (\partial)/(\partial u)(2sin(u)cos(v)) ?

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

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(\partial)/(\partial u)(2sin(u)cos(v))

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

2 cos (v) cos (u)

Solution steps

\frac{\partial}{\partial u} (2 sin (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} (sin (u))

Apply the common derivative:

\frac{\partial}{\partial u} (sin (u)) = cos (u)

= 2 cos (v) cos (u)

y = 4 (x + 1), y = 5 (x + 1), x = 7

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

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

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

\frac{d y}{d x} - 2 x y = e^{x^{2}}

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