What is (\partial)/(\partial x)(cos(xy)+sin(yz)) ?

The answer to (\partial)/(\partial x)(cos(xy)+sin(yz)) is -ysin(xy)

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(\partial)/(\partial x)(cos(xy)+sin(yz))

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

- y sin (x y)

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (sin (yz)) = 0

= - sin (x y) y + 0

Simplify

= - y sin (x y)

\frac{d}{d x} (\frac{8 x}{9 - tan ( x )})

\frac{- 5 x ^{2}}{2 x + 3}

\int sec^{- 2} (x t a) n^{3} x d x

\int_{1}^{7} π (x - 1) d x

\frac{d}{d x} (arcsin (\frac{a}{x}))

What is (\partial)/(\partial x)(cos(xy)+sin(yz)) ?
The answer to (\partial)/(\partial x)(cos(xy)+sin(yz)) is -ysin(xy)