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))