# d/dx

posted by
**Jen**
.

d/dx( ln |sin(pi/x)| ) = ?

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