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#