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

1 Answer
Aug 19, 2017

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!