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### 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

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

```
Associated Topics:
High School Calculus

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