Question

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

Answer 1

`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.

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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)`

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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)))`