What is (\partial)/(\partial y)(e^{-x}cos(yz)z) ?

The answer to (\partial)/(\partial y)(e^{-x}cos(yz)z) is -e^{-x}z^2sin(yz)

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(\partial)/(\partial y)(e^{-x}cos(yz)z)

\frac{\partial}{\partial y} (e^{- x} cos (yz) z)

- e^{- x} z^{2} sin (yz)

Solution steps

\frac{\partial}{\partial y} (e^{- x} cos (yz) z)

Treat

x, z

as constants

Take the constant out:

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

= e^{- x} z \frac{\partial}{\partial y} (cos (yz))

Apply the chain rule:

- sin (yz) \frac{\partial}{\partial y} (yz)

= - sin (yz) \frac{\partial}{\partial y} (yz)

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

= e^{- x} z (- sin (yz) z)

Simplify

e^{- x} z (- sin (yz) z) : - e^{- x} z^{2} sin (yz)

= - e^{- x} z^{2} sin (yz)

\frac{\partial}{\partial y} (3 x - 4 x y + 9 y)

\frac{d}{d x} (\frac{x - 2}{x - 1})

\int_{0}^{1} 5 x - 5 x^{2} d x

\frac{( s + 1 ) ^{2}}{s ^{2} - s - 6}

\frac{\partial}{\partial x} (sin (6 x^{3} y - 6 x y^{2}))

What is (\partial)/(\partial y)(e^{-x}cos(yz)z) ?
The answer to (\partial)/(\partial y)(e^{-x}cos(yz)z) is -e^{-x}z^2sin(yz)