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

The answer to (\partial)/(\partial y)(ycos(xy)+e^x) is cos(yx)-yxsin(yx)

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(\partial)/(\partial y)(ycos(xy)+e^x)

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (e^{x}) = 0

= cos (x y) - x y sin (x y) + 0

Simplify

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

\int cos^{2} (t) - sin^{2} (t) d t

\frac{\partial}{\partial x} (sin (x^{1} y^{3}))

(- 2 cos (2 x) - sin (3 x))^{2}

x \to \frac{1}{2} lim (arcsin (x))

What is (\partial)/(\partial y)(ycos(xy)+e^x) ?
The answer to (\partial)/(\partial y)(ycos(xy)+e^x) is cos(yx)-yxsin(yx)