Need help with logarithm simplification

Aug 2011
149
0
Hi

My task is to find derivative of
\(\displaystyle (\ln x)^x\)

so I got
\(\displaystyle (\ln x)^x \cdot(\ln(\ln x)+ \frac{1}{\ln(x)})\)

but in textbook it is simplified to
\(\displaystyle (\ln x)^x \cdot(\ln(x) +1)\)

my qouestion: could anyone show me steps to get \(\displaystyle \ln(x) +\)1 from previous step?

Edit: 1 more question:
(x^x)' was also simplified to x^x (ln x +1)

where does that +1 come from again?
 
Jul 2010
12,211
522
St. Augustine, FL., U.S.A.'s oldest city
1.) Let:

\(\displaystyle y=\(\ln(x)\)^x\)

Take the natural log of both sides:

\(\displaystyle \ln(y)=x\ln\(\ln(x)\)\)

Implicitly differentiate:

\(\displaystyle \frac{1}{y}\frac{dy}{dx}=x\frac{1}{x\ln(x)}+1\cdot\ln\(\ln(x)\)=\frac{1}{\ln(x)}+\ln\(\ln(x)\)\)

\(\displaystyle \frac{dy}{dx}=y\(\ln\(\ln(x)\)+\frac{1}{\ln(x)}\)\)

\(\displaystyle \frac{dy}{dx}=\(\ln(x)\)^x\(\ln\(\ln(x)\)+\frac{1}{\ln(x)}\)\)

You are right, the book is wrong. The two forms are not equivalent.
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Let ,

\(\displaystyle y \ = \ (\text{ln}x)^x\)

Then ,

\(\displaystyle \text{ln}y \ = \ x \text{ln}(\text{ln}x)\)

\(\displaystyle \frac{1}{y}y' \ = \ \cancel{x} \frac{1}{\text{ln}x} \frac{1}{\cancel{x}} \ + \ \text{ln}(\text{ln}x)\)

\(\displaystyle y' \ = \ (\text{ln}x)^x\( \frac{1}{\text{ln}x} \ + \ \text{ln}(\text{ln}x) \)\)

So unless we both have done some major fail the book is wrong.

:D
 
Jun 2013
1,315
116
London, England
You could have just checked on a spreadsheet or calculator that the two are not the same. Looks like a misprint: possibly copying the term ln(x) + 1 from the solution for (x^x)'.
 
Aug 2011
149
0
Thank you MarkFL, agentredlum, Pero.
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
You're welcome. This problem is tricky because one has to keep track of all the ln , in hindsight , an additional u substitution may be less confusing but one should proceed with caution regardless.

Let ,

\(\displaystyle y \ = \ ( \text{ln}x )^x\)

Let ,

\(\displaystyle u \ = \ \text{ln}x\)

\(\displaystyle u' \ = \ \frac{1}{x}\)

Then ,

\(\displaystyle y \ = \ u^x\)

\(\displaystyle \text{ln}y \ = \ x \text{ln}u\)

\(\displaystyle \frac{1}{y }y' \ = x \frac{1}{u}u' \ + \ \text{ln}u\)

Can you finish it?

:D