Using the Chain Rule: Finding a DerivativeDate: 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 |
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