Differentiate

ln(x)

----

..x

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- September 16th 2005, 08:20 AMcurtainsDifferentiation help please
Differentiate

ln(x)

----

..x - September 16th 2005, 10:50 AMticbol
ln(x)

----

..x

= ln(x) / x

Let y = ln(x) / x

Differentiate both sides with respect to x,

dy/dx = {x*[1/x] -ln(x) *[1]} / x^2

dy/dx = {1 -ln(x)} / x^2

That is it.

We used

d/dx (u/v) = {v*du -u*dv} / v^2 - September 17th 2005, 04:35 AMcurtains
Yes, but as I said in my original post (which seems to have been altered) x=e which changes things significantly, I got as far as you did but I can't get the last part to work out to the right answer.

- September 17th 2005, 04:50 AMticbol
x = e?

if so, then

ln(x) / x

= ln(e) / e

= 1/e

And differentiating that will give you zero because 1/e is a constant.

---------

If you substitute e for x in the derivative dy/dx = {1 -ln(x)} / x^2,

dy/dx = {1 -ln(e)} / e^2

dy/dx = {1 -1} / e^2

dy/dx = {0} / e^2

dy/dx = 0 ------------------still zero.