What is (\partial)/(\partial y)(sin(-2x)cos(8y)) ?

The answer to (\partial)/(\partial y)(sin(-2x)cos(8y)) is 8sin(2x)sin(8y)

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(\partial)/(\partial y)(sin(-2x)cos(8y))

\frac{\partial}{\partial y} (sin (- 2 x) cos (8 y))

8 sin (2 x) sin (8 y)

Solution steps

\frac{\partial}{\partial y} (sin (- 2 x) cos (8 y))

Simplify

sin (- 2 x) cos (8 y) : - sin (2 x) cos (8 y)

= \frac{\partial}{\partial y} (- sin (2 x) cos (8 y))

Treat

x

as a constant

Take the constant out:

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

= - sin (2 x) \frac{\partial}{\partial y} (cos (8 y))

Apply the chain rule:

- sin (8 y) \frac{\partial}{\partial y} (8 y)

= - sin (8 y) \frac{\partial}{\partial y} (8 y)

\frac{\partial}{\partial y} (8 y) = 8

= - sin (2 x) (- sin (8 y) \cdot 8)

Simplify

= 8 sin (2 x) sin (8 y)

(x + y) d x + x d y = 0

\int \frac{1}{5} cos (5 x) d x

x \to \infty lim (coth (x))

\int (2 x + 1) y^{2} sin (2 x y) d x

\frac{\partial}{\partial x} (x y sin (x) + y e^{x y})

What is (\partial)/(\partial y)(sin(-2x)cos(8y)) ?
The answer to (\partial)/(\partial y)(sin(-2x)cos(8y)) is 8sin(2x)sin(8y)