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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

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

f (x) = arctan (3 x)

\frac{s + 1}{s ^{2} + 4}

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

f^{^{'}} (x) = tan (x) + cot (x)

\int sin (a t) d t

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