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.

Emma Aldridge · Accepted Answer

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#

Isaiah Allen · Answer

undefined To prove quantitatively that for infinitesimally small $ \Delta x $, $ \frac{\Delta x}{x} $ is approximately equal to $ \Delta (\ln x) $, we'll use the definition of the derivative.

Consider the function $ y = \ln x $. The derivative of $ y $ with respect to $ x $ is given by:

$$ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} $$

Where $ \Delta y = \ln(x + \Delta x) - \ln(x) $ is the change in $ y $ corresponding to a small change $ \Delta x $ in $ x $.

Using the properties of logarithms, $ \Delta y = \ln\left(\frac{x + \Delta x}{x}\right) $.

Now, $ \frac{dy}{dx} = \lim_{\Delta x \to 0} \ln\left(\frac{x + \Delta x}{x}\right) $.

Recall that $ \ln(a) $ is approximately equal to $ a - 1 $ for small $ a $.

So, $ \frac{dy}{dx} = \lim_{\Delta x \to 0} \left(\frac{x + \Delta x}{x} - 1\right) $.

$$ = \lim_{\Delta x \to 0} \left(\frac{\Delta x}{x}\right) $$

Thus, for infinitesimally small $ \Delta x $, $ \frac{\Delta x}{x} $ is approximately equal to $ \Delta (\ln x) $.