Differentiation of ln(x)

Shaybay92 · May 29, 2009

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

I am unsure how to differentiate ln(x).

2. Relevant equations

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

3. The attempt at a solution

I let u = log_ex

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?

rock.freak667 · May 29, 2009

if y=ln(x) then, x=e^y

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

Shaybay92 · May 29, 2009

Thanks so much!

Shaybay92 · May 29, 2009

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

Shaybay92 said: ↑

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

I am unsure how to differentiate ln(x).

2. Relevant equations

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

3. The attempt at a solution

I let u = log_ex

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?

Cyosis · May 29, 2009

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]

Log in or Sign up

Differentiation of ln(x)

Shaybay92

rock.freak667

Shaybay92

Shaybay92

Cyosis

Differentiate ln ln ln x^3 (Replies: 1)

Differentiate (x^2)^(ln[x]) (Replies: 8)

Differentiate inverse (e^x + ln x ) (Replies: 14)

Differentiating ln(2^x+3^x) (Replies: 14)

Differentiate: ln(x) = yln(a) (Replies: 3)