# Differentiation of ln(x)

1. May 29, 2009

### Shaybay92

1. The problem statement, all variables and given/known data

I am unsure how to differentiate ln(x).

2. Relevant equations

$$\int$$ dx/ (x logex)

3. The attempt at a solution

I let u = logex

So it became:
$$\int$$ x-1u-1dx

To integrate I now need to find du/dx... which means differentiate ln(x). How does this work out?

2. May 29, 2009

### rock.freak667

if y=ln(x) then, x=ey

find dx/dy. Invert to get dy/dx and then figure out what eln[f(x)] works out to be.

3. May 29, 2009

### Shaybay92

Thanks so much!

4. May 29, 2009

### Shaybay92

Thanks but now I need help on the rest of the question! I'm really stuck. As, when I change it to integral f(x) du, the differentiated ln(x) does not cancel anything out.... Does anyone know how to integrate this equation?

$$\int$$ x-1u-1du/e^u

5. May 29, 2009

### Cyosis

Nothing "cancels" because you're not differentiating log(x) correctly. I don't really know how you can encounter these kind of problems without ever having seen the derivative of log(x), but this is how it works.

$$y=\log x \Rightarrow x=e^y, \;\; \frac{dx}{dx}=\frac{d e^y}{dx}=e^y \frac{dy}{dx}=1 \Rightarrow \frac{dy}{dx}=\frac{1}{e^y} \Rightarrow \frac{d log x}{dx}=\frac{1}{x}$$