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

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

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(\partial)/(\partial x)(cos(2x+4y))

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

- sin (2 x + 4 y) \cdot 2

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial x} (2 x + 4 y) = 2

= - sin (2 x + 4 y) \cdot 2

\frac{d}{d x} (- \frac{x ^{2}}{2})

3 \cdot y^{^{'}} = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}, y (1) = 3

f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 0

\int \frac{1}{x} - \frac{3}{x ^{2} + 1} d x

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