Question

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

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

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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.

Answer 1

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

Explanation:

`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`