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

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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#