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

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

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

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

2 sin (u) cos (v)

Solution steps

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

Treat

u

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= 2 sin (u) cos (v)

\frac{d}{d x} (\frac{5 x}{x + 2})

\int 5 e^{x} - \frac{1}{x} d x

sin (2 arccos (x))

\int_{0}^{\frac{π}{18}} cos (6 x) d x

\int_{0}^{x} 0

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