User:WindOwl/ln

The goal is to show that $$\frac{d}{dx}\ln \left( x \right)=\frac{1}{x}$$

Using the definition of e: $$e\equiv \underset{n\to 0}{\mathop{\lim }}\,\left( 1+n \right)^{1/n}$$

This is done using the definition of a derivative: $${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( x+h \right)+f\left( x \right)}{h}$$

Plugging in ln: $${f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\ln \left( x+h \right)+\ln \left( x \right)}{h}$$

Apply of the subtraction property of logarithms, then simplify it a bit: $$\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}\ln \left( \frac{x+h}{x} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{h}\ln \left( 1+\frac{h}{x} \right) \end{align}$$

This step is somewhat insane: (1/h) is the same as (x/h) * (1/x) right? Keeping this in mind, and apply another property of logs: $$\begin{align} & {f}'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\left( \frac{1}{x} \right)\left( \frac{x}{h} \right)\ln \left( 1+\frac{h}{x} \right) \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{x}\ln \left[ \left( 1+\frac{h}{x} \right)^{x/h} \right] \end{align}$$

Now, remember, h is the thing that’s changing here. The limit doesn’t affect that 1/x at all, so we can actually take it out. $${f}'\left( x \right)=\frac{1}{x}\underset{h\to 0}{\mathop{\lim }}\,\ln \left[ \left( 1+\frac{h}{x} \right)^{x/h} \right]$$

Now we use the definition of e above. The x in the equation doesn’t affect anything, because it’s constant with respect the h. The form is all the matters. As h gets closer and closer to zero, the expression inside the brackets gets closer and closer to e. The value of x probably affects how fast it gets there, but it’s still going to e.  Therefore: $$\begin{align} & {f}'\left( x \right)=\frac{1}{x}\ln \left( e \right) \\ & =\frac{1}{x} \end{align}$$