Prove quantitatively that for infinitesimally small $Deltax$ , $(Deltax)/x ~~ Delta(lnx)$ ?

Question

I actually have probably proved this, but I think I did it qualitatively. Not sure if it's what my book is looking for...

For some infinitesimally small $Deltax$ , supposedly, $(Deltax)/x ~~ Deltalnx$ . But if $Deltax$ is small, then $Deltax = dx$ , the differential change in $x$ .

That is, $1/xdx = d(lnx)$ . Integrating both sides:

$int 1/xdx = intd(lnx)dx$

The integral of a derivative cancels out to give:

$int 1/xdx = color(blue)(ln|x| + C)$

which we know to be true from calculus.

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Answer 1 · 2016-08-23T07:11:19

slightly different way of looking at it, but same idea.

$lim_(Delta x to 0) (Delta (ln x))/(Delta x) = (d(ln x))/(dx) = 1/x$ .

$therefore (Delta (ln x))/(Delta x) approx 1/x$

And so $Delta (ln x) approx (Delta x)/x$

or you could go more formal and write it as

$lim_(Delta x to 0) (Delta (ln x))/(Delta x) = lim_(Delta x to 0) (ln (x + Delta x) - ln x)/(Deltax)$

...and complete the derivation of the derivative of ln x from first principles.

So
$= lim_(Delta x to 0) 1/(Delta x) ln( (x + Delta x)/ x)$

$= lim_(Delta x to 0) 1/(Delta x) ln( 1 + (Delta x)/ x)$

$y = (Delta x )/ x$

$= lim_(y to 0) 1/(y x) ln( 1 + y)$

$= lim_(y to 0) 1/( x) ln( 1 + y)^(1/y)$

$= 1/( x) ln( e) = 1/x$