How do you find the derivative of $sin^2(lnx)$ ?

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Answer 1 · 2015-11-01T16:32:51

Recall that $sin^2(lnx) = [sin(lnx)]^2$ and use the chain rule twice.

$d/dx(sin^2(lnx)) =d/dx( [sin(lnx)]^2)$

$= 2sin(lnx) [d/dx(sin(lnx))]$

$= 2sin(lnx) [cos(lnx)d/dx(lnx)]$

$= 2sin(lnx) cos(lnx)[1/x]$

$= (2sin(lnx) cos(lnx))/x$

Which we may prefer to write as:

$= sin(2lnx)/x$

Or, perhaps as

$= sin(ln(x^2))/x$

How do you find the derivative of sin^2(lnx)sin^2(lnx)?