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.