Here we shall make use of the product rule, which states that d/(dx) (f(x)g(x)) = (df)/(dx)g(x) + f(x)(dg)/(dx)
= f'(x)g(x) + f(x)g'(x)
We will begin by finding the first derivative.
*Note: The below assumes the student is comfortable with the chain rule, the trigonometric identities, and the trigonometric derivatives. *
Here, f(x) = sin(ax+b) and g(x) = cos(ax+b). Via use of the chain rule and the definitions for trigonometric derivatives, this yields f'(x) = a cos(ax+b) and g'(x) = -a sin(ax+b). Thus, our derivative is
f'(x)g(x) + f(x)g'(x) = a cos(ax+b)cos(ax+b) +(-1)a sin(ax+b)sin(ax+b)
= a (cos^2 (ax+b) - sin^2(ax+b))
We know from the double angle identities that cos(2u) = (cos u)^2 - (sin u)^2 = cos^2 (u) - sin^2(u). Thus, for u = ax+b...
f'(x)g(x) + f(x)g'(x) = a (cos^2 u - sin^2 u) = a cos (2u) = a cos (2ax+2b)
This only yields the first derivative; however, from here, determining the later derivatives is a matter of recalling the angle-sum equations, specifically cos(a+b) = cos(a)cos(b) - sin(a)sin(b). For h(x) = cos(x), h'(x) = -sin(x), h"(x) = -cos(x), h"'(x) = sin(x), h""(x) = cos(x), and so forth.
Using our sum identity, we determine that cos(pi/2 +x) = -sin(x), cos(pi +x) = -cos(x), cos(3/2 pi +x) = sin (x), and cos(2pi+x) = cos(x) Thus, the n^(th) derivative of cos(x) = h^((n))x = cos((npi)/2 +x) In our case, however, we have h(x) = a cos(2ax+2b), and thus - via the Chain Rule - our n^(th) derivative would actually be h^((n))x = 2^n a^(n+1) cos((npi)/2+x)
Note also, however, that the n^(th) derivative of h(x) is actually the (n+1)^(th) derivative of our initial function. To find the n^(th) derivative of the initial function, we need to find the (n-1)^(th) derivative for h(x) Thus, for our initial function f(x)g(x) = j(x)...
j^((n))(x) = 2^(n-1)a^n cos(((n-1)pi)/2+x)
