Differentiation of ln(x)

[Shaybay92]

Homework Statement

I am unsure how to differentiate ln(x).

Homework Equations

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

The Attempt at a Solution

I let u = log_ex

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=e^y

find dx/dy. Invert to get dy/dx and then figure out what e^ln[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

Homework Statement

I am unsure how to differentiate ln(x).

Homework Equations

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

The Attempt at a Solution

I let u = log_ex

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.

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