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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = cos (y - 2 x)

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

\frac{\partial}{\partial y} (y) = 1

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

= 1 \cdot cos (y - 2 x) + (- sin (y - 2 x)) y

Simplify

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

\int y e^{2 y} d y

\int \frac{6 x ^{5} + 6}{( x ^{6} + 6 x ) ^{2}} d x

n = 0 \sum \infty 2 e^{- n}

\frac{\partial}{\partial x} (9 - x^{2})

\frac{d}{d x} (\frac{( 3 x ^{2} )}{2 x + 4})^{3}

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