How do you differentiate #f(x)= x^2/ (lnx)# twice using the quotient rule?

The quotient rule states that

#d/dx(g(x)/(h(x)))=(h(x)g'(x)-g(x)h'(x))/(h(x))^2#

So, for #f(x)#, we see that

#g(x)=x^2#
#h(x)=lnx#

Differentiate both of these:

#g'(x)=2x#
#h'(x)=1/x#

Plugging these both in, we see that

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

Which simplifies to be

#f'(x)=(2xlnx-x)/ln^2x#

We can apply the quotient rule to this again to find the second derivative. We will redefine our functions being divided:

#g(x)=2xlnx-x#
#h(x)=ln^2x#

However, this time, finding #g'(x)# and #h'(x)# will not be so straightforward.

The #2xlnx# term of #g(x)# will require the product rule to differentiate.

We see that

#g'(x)=lnxd/dx(2x)+2xd/dx(lnx)-1#

#g'(x)=lnx(2)+2x(1/x)-1#

#g'(x)=2lnx+1#

Finding #h'(x)# will require use of the chain rule:

#h'(x)=2lnxd/dx(lnx)=2lnx(1/x)=(2lnx)/x#

Thus, we see that

#g(x)=2xlnx-x#
#h(x)=ln^2x#

#g'(x)=2lnx+1#
#h'(x)=(2lnx)/x#

Hence the second derivative equals

#f''(x)=(ln^2x(2lnx+1)-(2xlnx-x)(2lnx)/x)/ln^4x#

Expanding completely, this equals

#f''(x)=(2ln^3x+ln^2x-4ln^2x+2lnx)/ln^4x#

Combining the #ln^2x# terms and dividing a common #lnx# term from the fraction, we see that

#f''(x)=(2ln^2x-3lnx+2)/ln^3x#