What is (\partial)/(\partial x)(-ae^{kz}wsin(kx-wt)) ?

The answer to (\partial)/(\partial x)(-ae^{kz}wsin(kx-wt)) is -ae^{kz}wcos(kx-wt)k

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(\partial)/(\partial x)(-ae^{kz}wsin(kx-wt))

\frac{\partial}{\partial x} (- a e^{k z} w sin (k x - wt))

- a e^{k z} w cos (k x - wt) k

Solution steps

\frac{\partial}{\partial x} (- a e^{k z} w sin (k x - wt))

Treat

a, k, z, w, t

as constants

Take the constant out:

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

= - a e^{k z} w \frac{\partial}{\partial x} (sin (k x - wt))

Apply the chain rule:

cos (k x - wt) \frac{\partial}{\partial x} (k x - wt)

= cos (k x - wt) \frac{\partial}{\partial x} (k x - wt)

\frac{\partial}{\partial x} (k x - wt) = k

= - a e^{k z} w cos (k x - wt) k

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What is (\partial)/(\partial x)(-ae^{kz}wsin(kx-wt)) ?
The answer to (\partial)/(\partial x)(-ae^{kz}wsin(kx-wt)) is -ae^{kz}wcos(kx-wt)k