What is (\partial)/(\partial y)(sin(xy)+xycos(xy)) ?

The answer to (\partial)/(\partial y)(sin(xy)+xycos(xy)) is 2xcos(xy)-x^2ysin(xy)

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(\partial)/(\partial y)(sin(xy)+xycos(xy))

\frac{\partial}{\partial y} (sin (x y) + x y cos (x y))

2 x cos (x y) - x^{2} y sin (x y)

Solution steps

\frac{\partial}{\partial y} (sin (x y) + x y cos (x y))

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

Simplify

cos (x y) x + x (cos (x y) - x y sin (x y)) : 2 x cos (x y) - x^{2} y sin (x y)

= 2 x cos (x y) - x^{2} y sin (x y)

What is (\partial)/(\partial y)(sin(xy)+xycos(xy)) ?
The answer to (\partial)/(\partial y)(sin(xy)+xycos(xy)) is 2xcos(xy)-x^2ysin(xy)