What is the derivative of f(x)=(lnx)^2/x^2?

1 Answer
Oct 19, 2016

f'(x) = (2lnx( 1-lnx ) )/ x^3

Explanation:

We can rewrite f(x)=(lnx)^2/x^2 as f(x) = (lnx/x)^2

Now, let u=lnx/x Then using the product rule:

d/dx(u/v)=( v(du)/dx-u(dv)/dx ) / v^2

So (du)/dx = ( xd/dx(lnx)-(lnx)d/dx(x) ) / x^2
:. (du)/dx = ( x1/x-(lnx)1 ) / x^2
:. (du)/dx = ( 1-lnx ) / x^2

And using the chain rule we now have:

f'(x) = d/dx{(lnx/x)^2 }
:. f'(x) = d/dxu^2
:. f'(x) = d/(du)u^2 xx (du)/dx
:. f'(x) = 2u (du)/dx

And substituting u and (du)/dx gives:
:. f'(x) = 2lnx/x xx ( 1-lnx ) / x^2
:. f'(x) = (2lnx( 1-lnx ) )/ x^3