What is (\partial)/(\partial x)(ycos(xy)-ysin(xy)) ?

The answer to (\partial)/(\partial x)(ycos(xy)-ysin(xy)) is -y^2sin(yx)-y^2cos(yx)

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(\partial)/(\partial x)(ycos(xy)-ysin(xy))

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (y sin (x y)) = y^{2} cos (y x)

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

h \to 0 lim (\frac{( sin ( \frac{π}{6} + h ) - \frac{1}{2} )}{h})

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

x \to 3 lim (x^{2} + 7)

\frac{d}{d x} (lo g_{2} (8^{x - 5}))

(- 11, - 4), (- 10, 5)

What is (\partial)/(\partial x)(ycos(xy)-ysin(xy)) ?
The answer to (\partial)/(\partial x)(ycos(xy)-ysin(xy)) is -y^2sin(yx)-y^2cos(yx)