What is (\partial)/(\partial y)(y+ycos(xy)) ?

The answer to (\partial)/(\partial y)(y+ycos(xy)) is 1+cos(xy)-xysin(xy)

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(\partial)/(\partial y)(y+ycos(xy))

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 1 + cos (x y) - x y sin (x y)

\frac{d r}{d x} = 4 π r^{2}

\frac{1}{2 x + 25}

\int \frac{19}{x 1 + 2 x} d x

lo g_{4} (e^{2 x})

\frac{d}{d x} (arccos (- \frac{x}{3}))

What is (\partial)/(\partial y)(y+ycos(xy)) ?
The answer to (\partial)/(\partial y)(y+ycos(xy)) is 1+cos(xy)-xysin(xy)