What is (\partial)/(\partial z)(2e^xcos(yz)) ?

The answer to (\partial)/(\partial z)(2e^xcos(yz)) is -2e^xysin(yz)

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(\partial)/(\partial z)(2e^xcos(yz))

\frac{\partial}{\partial z} (2 e^{x} cos (yz))

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

Solution steps

\frac{\partial}{\partial z} (2 e^{x} cos (yz))

Treat

x, y

as constants

Take the constant out:

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

= 2 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

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

Simplify

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

x \to 0 lim (x^{2} - x)

\frac{d}{d x} ((6 x)^{3 x})

\int sec (x) tan (x) d x

y^{^{''}} + 5 y^{^{'}} = (4 t - 2) sin (2 t)

\frac{d}{d t} (3 t e^{t})

What is (\partial)/(\partial z)(2e^xcos(yz)) ?
The answer to (\partial)/(\partial z)(2e^xcos(yz)) is -2e^xysin(yz)