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Differentiation of ln(x)

  • Thread starter Shaybay92
  • Start date
  • #1
124
0

Homework Statement



I am unsure how to differentiate ln(x).

Homework Equations



[tex]\int[/tex] dx/ (x logex)

The Attempt at a Solution



I let u = logex

So it became:
[tex]\int[/tex] x-1u-1dx

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

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
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
124
0
Thanks so much!
 
  • #4
124
0
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?

[tex]\int[/tex] x-1u-1du/e^u


Homework Statement



I am unsure how to differentiate ln(x).

Homework Equations



[tex]\int[/tex] dx/ (x logex)

The Attempt at a Solution



I let u = logex

So it became:
[tex]\int[/tex] x-1u-1dx

To integrate I now need to find du/dx... which means differentiate ln(x). How does this work out?
 
  • #5
Cyosis
Homework Helper
1,495
0
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.

[tex]
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}[/tex]
 

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