What are the first and second derivatives of # g(x) =x^2/(lnx)^2#?

1 Answer
Nov 21, 2015

#dy/dx = 2x*(ln(x)-1)/ln^3(x)#
#(dy^2)/dx^2 = (2 (ln^2(x)-3 ln(x)+3))/(ln^4(x))#

Explanation:

We have

#y = x^2/(ln(x))^2#

We can use the product rule, the quotient rule, or the chain rule with logs. To make our life easier I'll use log derivativing.

Take the log of both sides

#ln(y) = ln(x^2/(ln(x))^2)#

Use log properties

#ln(y) = 2ln(x) - 2ln(ln(x))#

Derivate implicity

#1/y*dy/dx = 2/x - 2/ln(x)*1/x#

Multiply by #y#

#dy/dx = (2x)/ln^2(x) - (2x)/ln^3(x)#

Or

#dy/dx = 2x*(ln(x)-1)/ln^3(x)#

To derivate again, do the same thing, now skipping over some steps

#ln(dy/dx) = ln(2x) + ln(ln(x)-1) - 3ln(ln(x))#

#dx/dy * (d^2y)/dx^2 = 2/(2x) + 1/(ln(x)-1)*1/x - 3/ln(x)*1/x#

Multiply both sides by #dy/dx#

#(dy^2)/dx^2 = 2*(ln(x)-1)/(ln^3(x)) + (2x)/ln^3(x) - 6*(ln(x)-1)/ln^4(x)#

Or, using algebra to tie it all together

#(dy^2)/dx^2 = (2 (ln^2(x)-3 ln(x)+3))/(ln^4(x))#