Using the Chain Rule: Finding a Derivative

Date: 2/26/96 at 8:47:29
From: Muhammad Momin Rashid
Subject: Derivative

I would like to know the solution to the following:

 d/dx sin(sin (2cos(2x/3))) = ?

Date: 2/28/96 at 11:11:31
From: Doctor Byron
Subject: Re: Derivative

Hi Muhammad,

Finding a derivative like this relies on the use of the 
chain rule:

d/dx(f(g(x)) = g'(x) * f'(g(x))

To apply your problem, we let f(x) = sin(x)
and g(x) = sin(2cos(2x/3))

d/dx(f(g(x)) = g'(x) * f'(g(x)) = d/dx sin(2cos(2x/3)) * cos(sin(2cos(2x/3))

Now we need to expand the first term using the chain rule.  
Apply the chain rule to d/dx sin(2cos(2x/3)) by letting 
h(x) = sin(x) and k(x) = 2cos(2x/3)

d/dx sin(2cos(2x/3)) = k'(x) * h'(k(x)) = d/dx 2cos(2x/3) * cos(2cos(2x/3)) 
= -4/3sin(2x/3) * cos(2cos(2x/3))

Substituting back into what we found before, we see that the 
derivative in question equals:

cos(sin(2cos(2x/3)) * cos(2cos(2x/3)) * -4/3sin(2x/3)

I hope this was helpful.

-Doctor Byron,  The Math Forum