Is it circular to use L'Hospital's rule to find derivative? I was killing time by doing some random math on paper and came up with the idea of trying to use L'Hospital on the limit definition of the derivative. If I choose the function $f(x)=\ln x$ then I can do
$$f'(x)=\lim_{h\to 0}\frac{\ln(x+h) - \ln x}{h}=\lim_{h\to 0}\frac{\ln(\frac{x+h}{x})}{h}=\lim_{h\to 0}\frac{\ln(1 + \frac hx)}{h}$$ The numerator approaches $\ln 1=0$ and the denominator approaches $0$, so I take the derivative of both with respect to $h$.
$$f'(x)=\lim_{h\to 0}\frac{f'(1 + \frac hx)\cdot \frac 1x}{1}=
\frac 1x \cdot \lim_{h\to 0}f'(1+\frac hx)$$
$$f'(x)=\frac{f'(1)}{x}$$ This just so happens to match the truth, but I suspect that this is a flawed argument. Can you tell me specifically where any holes in the logic might be?
 A: The argument is not circular, but it is incomplete.  We need to assume the following properties of $f(x) = \log x$:

*

*$f$ is differentiable in a neighborhood of $1$.

*$f$ obeys the property $f(a) + f(b) = f(ab)$ for all $a, b > 0$.

*$f$ permits $\lim_{x \to a} f'(g(x)) = f'(\lim_{x \to a}g(x))$; that is, the derivative of $f$ may be interchanged with the limit for a suitable function $g$.

Then,
$$\begin{align*}
f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} &&\text{(definition)} \\
&= \lim_{h \to 0} \frac{f(1 + \frac{h}{x})}{h} && \text{(property 2)} \\
&= \lim_{h \to 0} \frac{\frac{d}{dh}[f(1 + \frac{h}{x})]}{1} && \text{(property 1 + L'Hopital's rule)} \\
&= \lim_{h \to 0} f'\left(1 + \frac{h}{x}\right) \cdot \frac{d}{dh}\left[1 + \frac{h}{x}\right] && \text{(chain rule)} \\
&= \frac{1}{x} \lim_{h \to 0} f'\left(1 + \frac{h}{x}\right) \\
&= \frac{1}{x} f'\left( \lim_{h \to 0} 1 + \frac{h}{x} \right) && \text{(property 3)} \\
& \frac{f'(1)}{x}.
\end{align*}$$
Then this calculation shows that $f'(x)$ is proportional to some constant times $1/x$, and that this constant is the value of the derivative at $1$.  So we are not done; we need to ascertain this value.
A: Well, it's rather useless because you'll use the derivative you want to find.
$$\lim_{h\rightarrow 0}\frac{\ln(1+\frac{h}{x})}{h}=\lim_{h\rightarrow 0}\frac{\dfrac{1}{x}\times\dfrac{1}{1+\frac{h}{x}}}{1}=\lim_{h\rightarrow 0}\frac{1}{x}\times\frac{x}{x+h}=\frac{1}{x}.$$
If you want to calculate the derivative of $\ln$ using the formula, you can try the idea
$$\lim_{h\rightarrow 0}\frac{1}{h}\ln\left(1+\frac{h}{x}\right)=\lim_{h\rightarrow 0}\ln\left(\left(1+\frac{h}{x}\right)^\frac{1}{h}\right)$$
I guess you can continue from here.
A: The holes in your logic is that you are using things that you have been given to prove. Consider a simple example below which is not very different from yours ;
Prove that Elephant $= 1$
Proof : Elephant = 1
Hence, Elephant$ –1 = 0$
Now adding $1$ both sides, we have ;
Elephant$ = 1$ . Hence proved.
Certainly there are non-circular ways to derive this.
