# Math - again

I have added MathJax to Drupal to write better math formulas.
It reminds me of the very first website I ever made back in 1995 with plain html, line by line. It's a whole lot of extra formatting, and no WYSIWYG here, but it works. To test out things I reproduce a simple high school math proof.

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Let's take the derivative of $\ln \left(x\right)$
Remember that the definition of a derivative is as follows:
$\displaystyle\displaystyle\lim _{x\rightarrow a}\dfrac {f\left( x\right) -f\left( a\right) }{x-a}=f'\left( a\right)$
In that case we can write
$\displaystyle\lim _{x\rightarrow a}\dfrac {\ln \left( x\right) -\ln \left( a\right) }{x-a}$

Because we know that
$\log \left( a\right) -\log \left( b\right) =\log \left( \dfrac {a}{b}\right)$ and $\log \left( a\right) ^{n}=n\log \left( a\right)$
we can write the equation as follows:
$\displaystyle\lim _{x\rightarrow a}\dfrac {\ln \left( \dfrac {x}{a}\right) }{x-a}$
which is the same as
$\displaystyle\lim _{x\rightarrow a}\dfrac {1}{x-a}.\ln \left( \dfrac {x}{a}\right)$
thus
$\displaystyle\lim _{x\rightarrow a}\ln \left( \dfrac {x}{a}\right) ^{\dfrac {1}{x-a}}$

If we substitute $x-a=p$ which implies $x=a+p$ we can continue a bit more civilized, only realizing that if
$x\rightarrow a$ then $p\rightarrow 0$
So the equation becomes
$\displaystyle\lim _{p\rightarrow 0}\ln \left( \dfrac {a+p}{a}\right) ^{\dfrac {1}{p}}$
and can be simplified to
$\displaystyle\lim _{p\rightarrow 0}\ln \left( 1+\dfrac {p}{a}\right) ^{\dfrac {1}{p}}$

This time we are stuck with $\dfrac {p}{a}$ which we solve by substitution again:
$\dfrac {p}{a}=\dfrac {1}{n}$
which means
$\dfrac {1}{p}=\dfrac {n}{a}=n\dfrac {1}{a}$

And if $p\rightarrow 0$ then $n\rightarrow \infty$

This leads us to
$\displaystyle\lim _{n\rightarrow \infty }\ln \left( 1+\dfrac {1}{n}\right) ^{n\cdot \dfrac {1}{a}}$
and of course
$\displaystyle\lim _{n\rightarrow \infty }\dfrac {1}{a}\ln \left( 1+\dfrac {1}{n}\right) ^{n}$

Some might have forgotten it but this contains the definition of $e$, namely
$e=\displaystyle\lim _{n\rightarrow \infty } \left( 1+\dfrac {1}{n}\right) ^{n}$

And with that we can simply conclude that we end up with
$\dfrac {1}{a}.\ln \left( e\right) =\dfrac {1}{a}$
therefore
$f'\left(x\right) =\dfrac {1}{x}$

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It seems to work. Back to the studio, then.