What is (\partial)/(\partial x)(-aksin(kx-wt)) ?

The answer to (\partial)/(\partial x)(-aksin(kx-wt)) is -ak^2cos(kx-wt)

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(\partial)/(\partial x)(-aksin(kx-wt))

\frac{\partial}{\partial x} (- ak sin (k x - wt))

- a k^{2} cos (k x - wt)

Solution steps

\frac{\partial}{\partial x} (- ak sin (k x - wt))

Treat

a, k, w, t

as constants

Take the constant out:

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

= - ak \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

= - ak cos (k x - wt) k

Simplify

- ak cos (k x - wt) k : - a k^{2} cos (k x - wt)

= - a k^{2} cos (k x - wt)

f (x) = e^{- 2.5 x^{2}}

n = 1 \sum \infty (- 1)^{n - 1} (\frac{7}{9 ^{n} - 1})

\frac{d}{d x} (\frac{x ^{4} + 9}{x ^{2}})

sin (e^{x})

What is (\partial)/(\partial x)(-aksin(kx-wt)) ?
The answer to (\partial)/(\partial x)(-aksin(kx-wt)) is -ak^2cos(kx-wt)