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

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

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

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

- cos (2 y) cos (2 x) \cdot 2

Solution steps

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

Simplify

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

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

cos (2 x) \frac{\partial}{\partial x} (2 x)

= cos (2 x) \frac{\partial}{\partial x} (2 x)

\frac{\partial}{\partial x} (2 x) = 2

= - cos (2 y) cos (2 x) \cdot 2

\frac{\partial}{\partial x} (2 x y e^{- x^{2} y})

f (x) = e^{x} \cdot cos (x), at x = 0

\int z^{m} d z

\frac{1}{1 - x}, 0

\int \frac{x}{1 - x} d x

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