Differentiation of ln(x)

[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?

[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.

[Shaybay92]

Thanks so much!

[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


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?

[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}