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))#
