What is the derivative of #ln(x)^x#?

1 Answer
Apr 21, 2015

We're going to use implicit differentiation to solve this problem.

We'll also be using the product rule and the chain rule .

Know that:

If #q=u*v# and #u=g(x)# and #v=h(x)#,

#(dq)/(dx)=u*(dv)/(dx)+v*(du)/(dx)#

So, say that:

#q=x*ln(ln(x))=u*v#

Therefore:

#u=x#, #(du)/(dx)=1#

Say that:

#v=ln(ln(x))=lnp#

Therefore:

#(dv)/(dp)=1/p=1/ln(x)#

#p=ln(x)#, #(dp)/(dx)=1/x#

This means that:

#(dv)/(dx)=1/(x*ln(x))#

And as a result...

#(dq)/(dx)=x*1/(x*ln(x))+ln(ln(x))*1#

#=1/(ln(x))+ln(ln(x))#

Now let's differentiate the function you were talking about using implicit differentiation...

#y=ln(x)^x#

#lny=ln(ln(x)^x)#

#lny=xln(ln(x))#

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

#y*1/y*(dy)/(dx)=y*{1/(ln(x))+ln(ln(x))}#

#(dy)/(dx)=y/(ln(x))+yln(ln(x))#

And finally...

#(dy)/(dx)=(ln(x)^x)/(ln(x))+ln(x)^x*ln(ln(x))#