We can define a function
L(x) as follows:
L(x)=∫1xt1dtwhich computes the area under
1/x from 1 to some arbitrary value
x. This function is, in fact, the same as
lnx.
1.
L(1)=0 since the area under
1/x from 1 to 1 is 0
2. It should be more-or-less obvious that if we differentiate
L(x) we get back to
1/t (since differentiating undoes integration)
3.
L(ab)=L(a)+L(b) so it behaves like a log function
That's because
L(ab)=∫1abt1dt=∫1at1dt+∫aabt1dt by splitting the area from 1 to
ab into 2 chunks and adding them.
Now the first integral
∫1at1dt is simply
L(a) by the defn of the function. Also if you make the substitution
x=u/a you can show that the second integral
∫aabt1dt is simply
L(b) as we wanted.
So
L(x) has all the properties of
lnx, and hence *is*
lnx and also (obviously?) differentiates to
1/x.
That is a more advanced way to answer the question but not inaccessible to a reasonably bright further maths student, I think.