#f(x)=(ln(x))^(secx)#
Here i'm writing #f(x)=y# for convenience.
Here we can see that a function is nested inside another function (#(lnx)^secx#), so we will take #color(red)(log)# on both sides.
#color(red)(log)y=color(red)(log)(lnx)^(secx)#
Using the logarithmic property of #logx^n=n logx# we can write this as,
#color(red)(log)y=secxcolor(red)(log)(lnx)#
Now we will differentiate both sides with respect to #x# using the chain rule and the product rule.
#color(violet)1.#The chain rule says that #->#
#d/dxU(V(x))=d/dxU(V(x)) xx d/dxV(x) xx d/dxx#
#color(violet)2.#The product rule says that #->#
#d/dxU*V=Ud/dxV+Vd/dxU#
Now back to the question #->#
#d/dx(logy)=d/dxsecx*log(lnx)#
#1/y d/dxy=log(lnx)d/dxsecx+secxd/dxlog(lnx)#
#1/y y'=log(lnx)*secx*tanx+secx*1/lnx*1/x#
Therefore,
#y'=y xx (log(lnx)*secx*tanx+secx*1/lnx*1/x)#
You can put the value of #y# (#f(x)#) in there if you want.
