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

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

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

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

- 2 sin (u) sin (v)

Solution steps

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

Treat

u

as a constant

Take the constant out:

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

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

Apply the common derivative:

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

= 2 sin (u) (- sin (v))

Simplify

= - 2 sin (u) sin (v)

e^{x}, \frac{e ^{2 x}}{2}, 0, 1

\frac{d}{d x} (- \frac{1}{4 ( x ) ^{\frac{3}{2}}})

\frac{4 x}{( x ^{2} + 1 ) ^{2}}

y = \frac{5 x ^{2} + 8 x + 4}{x}

y = sin (3 x + 4 y)

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