Integrating 1/x (Reciprocal Function)

Let's say we want to find a function $f(x)$ that satisfies $$\frac{\text df(x)}{dx}=f(x)$$ Or $$f(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

Then we might want to find an $f(x)$ that satisfies the functional equation $f(x+h)=f(x)f(h)$ so that we may "factor out" $f(x)$ and solve for $f(h)$.

As it turns out, the only solutions to that functional equation are $f(x)=0$, or $f(x)= A^x$.

Since we're not interested in the trivial case of $f(x)=0$, we will set $f(x)=e^x$ for some constant $e$.

$$e^x=\lim_{h\to0}\frac{e^xe^h-e^x}{h}$$ $$1=\lim_{h\to0}\frac{e^h-1}{h}$$

Putting this equation in terms of $A$, we see: $$\lim_{h\to0}(1+h)^{1/h}=e$$ Finally, if we substitute $n=\frac1h$: $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$$

Note that one can also use this to find the Taylor series of $e^x$ ($u=nx$): $$e^x=\lim_{n\to\infty}\left(1+\frac1n\right)^{nx}=\lim_{u\to\infty}\left(1+\frac{x}{u}\right)^{u}=\lim_{u\to\infty}\sum_{k=0}^u{u\choose k}\frac{x^n}{u^n}$$ Now, consider: $$\lim_{u\to\infty}\frac{u!}{(u-k)!u^k}=\lim_{u\to\infty}\prod_{i=0}^{k-1}\frac{u-i}{u}=\prod_{i=0}^{k-1}\left(\lim_{u\to\infty}\frac{u-i}{u}\right)=\prod_{i=0}^{k-1}1=1$$ Therefore $$e^x=\lim_{u\to\infty}\sum_{k=0}^u\frac{u!x^n}{(u-k)!k!u^k}=\sum_{k=0}^\infty\frac{x^k}{k!}$$

Now, consider the inverse function of $e^x$: $f^{-1}(x):=\ln x$.

We already know how to find the derivative of an inverse function, so let's apply that here:

$$\frac{\text d\ln x}{\text dx}=\frac{1}{e^{\ln x}}=\frac1x$$

That's right! $\ln x+C$ is the unique antiderivative of $x^{-1}$, which is important because the antiderivative of $x^n$ is generally $\frac{1}{n+1}x^{n+1}$, which clearly does not apply when $n=-1$.

The derivative of the logarithm can be useful in a variety of problem-solving settings, as $$\frac{\text d\ln(f(x))}{\text dx}=\frac{f'(x)}{f(x)}$$