What is (\partial)/(\partial x)(sin(-2x)cos(4y)) ?

The answer to (\partial)/(\partial x)(sin(-2x)cos(4y)) is -cos(4y)cos(2x)*2

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(\partial)/(\partial x)(sin(-2x)cos(4y))

\frac{\partial}{\partial x} (sin (- 2 x) cos (4 y))

- cos (4 y) cos (2 x) \cdot 2

Solution steps

\frac{\partial}{\partial x} (sin (- 2 x) cos (4 y))

Simplify

sin (- 2 x) cos (4 y) : - sin (2 x) cos (4 y)

= \frac{\partial}{\partial x} (- sin (2 x) cos (4 y))

Treat

y

as a constant

Take the constant out:

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

= - cos (4 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 (4 y) cos (2 x) \cdot 2

\frac{d}{d x} (x^{2} + a \cdot x + b)

\frac{d y}{d x} = x^{2} (1 + y)

\frac{d}{d x} (ln (\frac{x + 3}{x - 3}))

\int \frac{1}{x 8 - ln ^{2} ( x ) + 2 ln ( x )} d x

x \to 0 + lim (6 x ln (x))

What is (\partial)/(\partial x)(sin(-2x)cos(4y)) ?
The answer to (\partial)/(\partial x)(sin(-2x)cos(4y)) is -cos(4y)cos(2x)*2