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

The answer to (\partial)/(\partial x)(ycos(y-2x)) is 2ysin(y-2x)

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(\partial)/(\partial x)(ycos(y-2x))

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

2 y sin (y - 2 x)

Solution steps

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

Treat

y

as a constant

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= y (- sin (y - 2 x) (- 2))

Simplify

= 2 y sin (y - 2 x)

\int \frac{- y}{y} d y

y^{^{'}} + 2 x (y + 1) = 0

\int_{- 1}^{1} 2 x sin (1 - x^{2}) d x

x^{2 + 6 x}

\int \frac{cos ( 2 x ) - 3}{cos ( 2 x )} d x

What is (\partial)/(\partial x)(ycos(y-2x)) ?
The answer to (\partial)/(\partial x)(ycos(y-2x)) is 2ysin(y-2x)