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

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

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

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

2 sin (v) cos (u)

Solution steps

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

Treat

v

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= 2 sin (v) cos (u)

\int x^{2} sin (n x) d x

n = 0 \sum \infty (- 2)^{n}

\frac{\partial}{\partial x} (cos (4 y) e^{x})

x \to 2 lim (\frac{5 x - 25}{x - 2})

x^{^{'}} = - 0.0002 x

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