What is (\partial)/(\partial z)(4e^xcos(yz)) ?

The answer to (\partial)/(\partial z)(4e^xcos(yz)) is -4e^xysin(yz)

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(\partial)/(\partial z)(4e^xcos(yz))

\frac{\partial}{\partial z} (4 e^{x} cos (yz))

- 4 e^{x} y sin (yz)

Solution steps

\frac{\partial}{\partial z} (4 e^{x} cos (yz))

Treat

x, y

as constants

Take the constant out:

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

= 4 e^{x} \frac{\partial}{\partial z} (cos (yz))

Apply the chain rule:

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

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

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

= 4 e^{x} (- sin (yz) y)

Simplify

= - 4 e^{x} y sin (yz)

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

\int_{- \infty}^{2} \frac{1}{x ^{2} + 1} d x

\frac{\partial}{\partial y} (e^{x} + yz)

\int \frac{1}{e ^{z} + e ^{- z}} d z

\frac{d}{d x} ((3 + x - 2 x^{2})^{\frac{1}{4}})

What is (\partial)/(\partial z)(4e^xcos(yz)) ?
The answer to (\partial)/(\partial z)(4e^xcos(yz)) is -4e^xysin(yz)