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December 18, 2014

December 18, 2014

Posted by **Jen** on Wednesday, December 13, 2006 at 9:42pm.

Thanks.

If those are absolute value signs, the derivative will not exist when sin (pi/x) = 0, because of the sign change that occurs there. Assume sin (pi/x) > 0

Let u(x) = pi/x and v(x) = sin x, and use the chain rule.

d/dx ln v(u(x))=

d/dv dv/du du/dx

= -[1/(sin (pi/x])*cos x*(pi/x^2)

That assumes sin (pi/x) is positive. Change the sign if it is negative

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