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)

\frac{d}{d x} (\frac{5 x}{7 x ^{2} + 1})

y^{^{''}} + a y = 0

n = 1 \sum \infty n^{2}

\int_{- 8}^{8} 64 - x^{2} d x

\frac{d y}{d x} = 3 - 6 x + y - 2 x y

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