# Need help with logarithm simplification

#### rain

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?

#### MarkFL

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
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. #### Pero

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)'.

#### rain

Thank you MarkFL, agentredlum, Pero.

#### agentredlum

Math Team
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? 