What is the derivative of f(x) = (lnx)^(x)?

1 Answer
Nov 12, 2015

d/dx ln^x(x) = ln^x(x)(ln(ln(x))+1/ln(x))

Explanation:

We will be using several techniques in the evaluation of this derivative.


The product rule:
d/dxf(x)g(x) = f'(x)g(x) + f(x)g'(x)

The chain rule:
d/dxf(g(x)) = f'(g(x)g'(x)

Implicit differentiation:
d/dx f(y) = (df(y))/dydy/dx

Derivative of the natural logarithm:
d/dx ln(x) = 1/x

(a property of logarithms)
ln(x^n) = nln(x)


Now, we proceed to evaluate the derivative.
Let y = ln^x(x)

=> ln(y) = ln(ln^x(x)) = xln(ln(x)) (property of logarithms)

=> d/dxln(y) = d/dxxln(ln(x))

Through implicit differentiation and the derivative of natural log
d/dxln(y) = d/dyln(y)dy/dx = 1/ydy/dx

For the right hand side
d/dxxln(ln(x)) = 1*ln(ln(x)) + x(d/dxln(ln(x))) (product rule)
d/dxln(ln(x)) = 1/(ln(x))*1/x = 1/(xln(x)) (chain rule)
=> d/dxxln(ln(x)) =ln(ln(x)) + x/(xln(x))= ln(ln(x)) + 1/(ln(x))

So we have

1/ydy/dx = ln(ln(x)) + 1/(ln(x))

Multiplying by y gives us

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

Finally, we substitute back in for y to get

d/dxln^x(x) = ln^x(x)(ln(ln(x)) + 1/(ln(x)))