Derivative of ( ln ⁡ x ) <mrow class="MJX-TeXAtom-ORD">

Question

Derivative of

(\ln x)^{\ln x}

How can I differentiate the following function?

f (x) = (\ln x)^{\ln x} .

Is it a composition of functions? And if so, which functions?
Thank you.

Answer 1

The function

f

can be expressed as

f (x) = (\ln x)^{\ln x} = e^{(\ln x) \ln (\ln x)} .

Hence

f (x) = \exp (g (x))

, where

g (x) = (\ln x) \ln (\ln x)

, and thus

f^{'} (x) = {(e^{(\ln x) \ln (\ln x)})}^{'} = e^{(\ln x) \ln (\ln x)} {((\ln x) \ln (\ln x))}^{'} = (\ln x)^{\ln x} (\frac{1}{x} \ln (\ln x) + \ln x \frac{1}{x \ln x}) .

Answer 2

Note that

a^{b} = {(e^{\ln a})}^{b} = e^{b \ln a}

so

{(\ln x)}^{\ln x} = {(e^{\ln \ln x})}^{\ln x} = e^{\ln x \cdot \ln \ln x} = x^{\ln \ln x}

either of which which you should be able to do with methods you already know.
We can apply this technique generally to calculate the derivative of

f (x)^{g (x)}

:

f (x)^{g (x)} = e^{g \ln f}

so

\begin{aligned} \frac{d}{d x} f^{g} = \frac{d}{d x} e^{g \cdot \ln f} & = (\frac{d}{d x} (g \cdot \ln f)) e^{g \cdot \ln f} (chain rule) \\ = (\frac{d}{d x} (g \cdot \ln f)) f^{g} \\ = (g^{'} \ln f + \frac{g f^{'}}{f}) f^{g} (product rule) \end{aligned}

where

f^{'} = \frac{d f}{d x}

and

g^{'} = \frac{d g}{d x}

Answers (2)