Question

How do you find the second derivative of `y=Acos(Bx)`?

Answer 1

Well, start with the first derivative.
Treat it like the product of two functions: `f(x) * g(x)`
where `f(x) = A` (just a constant) and `g(x) = cos(Bx)`

so `(df(x))/dx = 0`.
and `(dg(x)/dx) = -Bsin(BX)` (used the chain rule here.)

So, by the product rule, the deriv. of the product `f(x) * g(x) =`

`((df(x))/dx) g(x) + f(x)((dg(x))/dx)`

in this case, that works out to `-ABsin(BX)`

Now, just do it again:

`d/dx(-ABsin(BX)) = -AB^2cos(BX)` (by the chain rule)

GOOD LUCK!