What is (\partial)/(\partial z)(9e^xcos(yz)) ?

The answer to (\partial)/(\partial z)(9e^xcos(yz)) is -9e^xysin(yz)

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(\partial)/(\partial z)(9e^xcos(yz))

\frac{\partial}{\partial z} (9 e^{x} cos (yz))

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

Solution steps

\frac{\partial}{\partial z} (9 e^{x} cos (yz))

Treat

x, y

as constants

Take the constant out:

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

= 9 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

= 9 e^{x} (- sin (yz) y)

Simplify

= - 9 e^{x} y sin (yz)

f (x) = x^{3} 2 x + 1

\int_{1}^{100} \frac{3}{x ^{3}} d x

x \to 6 lim (x^{- 1})

x - y^{2} = 1, x + y = 7

What is (\partial)/(\partial z)(9e^xcos(yz)) ?
The answer to (\partial)/(\partial z)(9e^xcos(yz)) is -9e^xysin(yz)