Let f(x)=(ln(x))^(sec(x)) . f'(x) =?

1 Answer
Nov 15, 2017

f'(x)=f(x) xx (log(lnx)*secx*tanx+secx*1/lnx*1/x)

Explanation:

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.