What is the derivative of $sin^2(lnx)$ ?

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Answer 1 · 2015-12-23T17:32:50

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

First, treat $sin^2(lnx)$ as $u^2$ where $u=sin(lnx)$ .

$d/dx[u^2]=2u*u'$

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

To find $d/dx[sin(lnx)]$ , treat it as $sin(v)$ where $v=lnx$ .

$d/dx[sin(v)]=cos(v)*v'$

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

Also, $d/dx[lnx]=1/x$ .

Plug this in:

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

Plug this in:

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

What is the derivative of sin^2(lnx)sin^2(lnx)?