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)#
