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

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

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

\frac{\partial}{\partial y} (y \cdot 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)

f (x) = x^{2} - x + 1

x \to 0 lim (x^{3})

\frac{d}{d x} ((3 x^{2} + 3 x - 1)^{4})

8 y^{^{''}} - 2 y^{^{'}} - y = 4 + 5 sin (2 t)

y = x^{2} - 3 x, y = 2 x + 6

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