Derivative Of $\ln(x)$

Question

It is required to find the derivative of the natural logarithm of $x$: $\frac {d}{dx}\ln(x)$
My solution:
Let $f(x)=\ln(x) $ then $f'(x)=\frac {d}{dx}\ln(x) $
By definition:$$f'(x)= \lim_{h\to 0}\frac{f(x+h)- f(x)}{h} $$ By substitution:$$f'(x) = \lim_{h\to 0}\frac{\ln(x+h)- \ln(x)}{h} $$ Since $\ln(a)-\ln(b)=\ln(\frac ab)$, then: $$f'(x) = \lim_{h\to 0}\frac{\ln(\frac{x+h}x)}{h}$$ Since $a\times \ln(b) = \ln(a^b)$, then: $$f'(x) = \lim_{h\to 0}\ln((\frac{x+h}x)^\frac 1h)$$ Then:$$f'(x) = \lim_{h\to 0}\ln((1+\frac hx)^\frac 1h)$$ How to continue?
Are there any other ways to find the derivative?
Thanks in advance!
Note: It is not allowed to use the fact that: $(e^x)'=e^x$
$$ e=\lim\limits_{n\to\infty} \left(1+\frac1n\right)^n$$

Answer 1

You're almost there. Recall the definition of $e$:

$e=\lim\limits_{n\to\infty} \left(1+\frac1n\right)^n$

Or equivalently:

$e=\lim\limits_{h\to 0^+} \left(1+h\right)^{1/h}$

You can use that

$e^u=\lim\limits_{h\to 0^+} \left(1+hu\right)^{1/h}$

Can you take it from here?

Answer 2

Hint: You're nearly there with your approach. One method is to use substitution and let $\frac{h}{x} = u$, then consider the limit as $u \rightarrow 0$.

Note that $\lim \limits_{u \rightarrow 0}(1+u)^{\frac{1}{u}} = e$ (Why?)

Answer 3

I usually do this via the chain rule: \begin{align} y & = \log x \qquad \text{(or ``$\ln x$'' if you like)} \\[10pt] x & = e^y \\[10pt] \frac{dx}{dy} & = e^y \\[10pt] \frac{dx}{dy} & = x \\[10pt] \frac{dy}{dx} & = \frac 1 x. \end{align} Where do you see the chain rule above? I wasn't explicit about that. Here's a more explicit version: \begin{align} x & = \exp(\log x). \\[10pt] 1 = \frac{dx}{dx} & = \underbrace{\frac d {dx} \exp\log x = (\exp'\log x) \cdot\log' x}_{\text{chain rule}} = (\exp\log x) \cdot \log' x = x\log'x. \\[10pt] 1 & = x \log' x. \\[10pt] \frac 1 x & = \log' x. \end{align}