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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = cos (x y)

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

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

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

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

Simplify

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

x \to \infty lim (\frac{14}{x})

(\frac{( x - 6 )}{( x + 7 )})^{3}

x \to 0 lim (ln (\frac{x}{1 - x}))

h (t) = (t + 3)^{\frac{2}{3}} (2 t^{2} - 4)^{3}

\int (e^{3 x} - 2)^{2} d x

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