Correcting Calculus Derivative: x(t) = A cos(ωt - \varphi)

[Agrasin]

I'm brushing up on calculus. I don't see how this derivative works.

x(t) = A cos(ωt - $\varphi$)

v(t) = dx / dt = - Aωsin(ωt + $\varphi$)



I get that the derivative of cosx is -sinx. I get that the omega is brought outside the cos function because of U substitution.

Why does the the minus sign turn into a plus sign?

Btw I saw this on wikipedia: http://en.wikipedia.org/wiki/Simple_harmonic_motion

 

[jbunniii]

Looks like a typo on the Wikipedia page. The correct derivative of $A \cos(\omega t - \varphi)$ is $-A\omega\sin(\omega t - \varphi)$.

Since $-\sin(x) = \sin(-x)$, an equivalent expression is $A\omega\sin(-\omega t + \varphi)$.

But $-A\omega\sin(\omega t + \varphi)$ is not equal to either of these, as can easily be seen by evaluating at $\omega = 0$, $\varphi = \pi/2$.