For this proof, we'll need the following background mathematics.

First principles formula for the derivative of a function `f(x)`, that is:

`(df)/(dx) = lim_{h->0}(f(x+h)-f(x))/h`

The logarithm laws

`log a - log b= log (a/b)`

and

`n log a = log a^n`.

Graph of `y=(1+t)^{1"/"t}`

The fact "log" and `e` are inverses, so

`log_e(e) = 1`.

The well-known limit

`lim_{t->0}(1+t)^{1"/"t} = e`

(We will not prove this limit here. The graph on the right demonstrates that as `t->0`, the graph of `y=(1+t)^{1"/"t}` approaches the value `e~~2.71828`.)

I will write `log(x)` to mean `log_e(x) = ln(x)`, to make it easier to read.

We have `f(x) = log(x)` so the derivative will be given by:

`(df)/(dx) = lim_{h->0}(log(x+h)-log(x))/h `

Now the top of our fraction is

`log(x+h)-log(x)` ` = log((x+h)/x)` ` = log(1 + h/x)`.

To simplify the algebra, we now substitute `t=h/x` and this gives us `h = xt`. Of course,

`lim_{h->0}(h) = lim_{t->0}(xt) = 0`.

So we have now:

`(log(x+h)-log(x))/h` ` = 1/(xt)log(1 + t)`.

We write this as:

`1/x[1/tlog(1 + t)]`.

Considering the expression in square brackets, we use the log law

`n log a = log a^n`

and write:

`1/tlog(1 + t) = log(1+t)^(1"/"t}`

Next, the limit of the expression `(1+t)^(1"/"t}` is:

`lim_{t->0}(1 + t)^{1"/"t} = e`.

Because `log(e)=1`, we can conclude:

`(dlog(x))/(dx) = lim_{h->0}(log(x+h)-log(x))/h`

`= 1/x log(lim_{t->0}(1 + t)^{1"/"t})`

` = 1/x log(e)`

` = 1/x`

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