What is (\partial)/(\partial z)(-3e^xcos(yz)) ?

The answer to (\partial)/(\partial z)(-3e^xcos(yz)) is 3e^xysin(yz)

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(\partial)/(\partial z)(-3e^xcos(yz))

\frac{\partial}{\partial z} (- 3 e^{x} cos (yz))

3 e^{x} y sin (yz)

Solution steps

\frac{\partial}{\partial z} (- 3 e^{x} cos (yz))

Treat

x, y

as constants

Take the constant out:

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

= - 3 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

= - 3 e^{x} (- sin (yz) y)

Simplify

= 3 e^{x} y sin (yz)

\int \frac{5 + 3 x}{1 + x ^{2}} d x

\int 15 x + 1 d x

e^{x} y \frac{d y}{d x} = 1

y = \frac{2}{x}, (- 2, - 1)

\int \frac{x}{2 x ^{3}} d x

What is (\partial)/(\partial z)(-3e^xcos(yz)) ?
The answer to (\partial)/(\partial z)(-3e^xcos(yz)) is 3e^xysin(yz)