What is (\partial)/(\partial z)(xycos(x+yz)) ?

The answer to (\partial)/(\partial z)(xycos(x+yz)) is -xy^2sin(x+yz)

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(\partial)/(\partial z)(xycos(x+yz))

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

- sin (x + yz) \frac{\partial}{\partial z} (x + yz)

= - sin (x + yz) \frac{\partial}{\partial z} (x + yz)

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

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

Simplify

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

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

What is (\partial)/(\partial z)(xycos(x+yz)) ?
The answer to (\partial)/(\partial z)(xycos(x+yz)) is -xy^2sin(x+yz)