Derivative of ln(x): Definition & Calculation

[Amer]

Is it possible to find the derivative of ln(x) by the definition how ?

$$f'(x) = \lim _ {h\rightarrow 0 } \frac{f(h+x) - f(x)}{h}$$

 

[Reckoner]

Is it possible to find the derivative of ln(x) by the definition how ?

See http://www.mathhelpboards.com/f35/problem-week-14-july-2nd-2012-a-1343/#post6686.

 

[Siron]

Here is a possible proof.
To prove: $$\lim_{h \to 0} \frac{\ln(x+h)-\ln(x)}{h} = \frac{1}{x}$$
Proof:
$$\lim_{h \to 0} \frac{\ln(x+h)-\ln(x)}{h} = \lim_{h \to 0} \frac{\ln\left(\frac{x+h}{x}\right)}{h} = \lim_{h \to 0} \frac{\ln\left(1+\frac{h}{x}\right)}{h}$$

We can use the maclaurin serie of $\ln(1+x)$ now.
$$\ln\left(1+\frac{h}{x}\right) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}\left(\frac{h}{x}\right)^{n+1}$$

Thus
$$\lim_{h \to 0} \frac{\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}\left(\frac{h}{x}\right)^{n+1}}{h}$$
$$=\lim_{h \to 0} \sum_{n=0}^{\infty} \frac{(-1)^n}{n+1}\frac{h^n}{x^{n+1}}$$
$$= \lim_{h \to 0} \left[\frac{1}{x} + \sum_{n=1}^{\infty} \frac{(-1)^n}{n+1}\frac{h^n}{x^{n+1}}\right]$$
$$= \frac{1}{x} + \sum_{n=1}^{\infty} \left[ \lim_{h \to 0} \frac{(-1)^n}{n+1}\frac{h^n}{x^{n+1}}\right] = \frac{1}{x}$$

 

[soroban]

Hello, Amer!

Is it possible to find the derivative of $$f(x) \,=\,\ln x$$ by the definition? .How?

$$f'(x)\:=\:\lim_{h\to0}\frac{f(x+h)- f(x)}{h}$$

$$f(x+h) - f(x) \;=\; \ln(x+h) - \ln(x) \;=\;\ln\left(\frac{x+h}{x}\right) \;=\;\ln\left(1 + \frac{h}{x}\right)$$$$\frac{f(x+h) - f(x)}{h} \;=\;\frac{1}{h}\ln\left(1 + \frac{h}{x}\right) \;=\;\ln\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}$$

. . . . . . . . . . . . $$=\;\ln\left(1 + \frac{h}{x}\right)^{\frac{1}{h}\cdot\frac{x}{x}} \;=\; \ln\left[\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}$$$$f'(x) \;=\;\lim_{h\to0}\frac{f(x+h) - f(x)}{h} \;=\;\lim_{h\to0}\left[\ln\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}$$

Let $$u \,=\,\frac{x}{h}$$ . . Note: if $$h\to0$$, then $$u \to\infty$$
We have: .$$\lim_{u\to\infty}\left[\ln\left(1 + \frac{1}{u}\right)^u\right]^{\frac{1}{x}} \;=\;\ln\left[\underbrace{\lim_{u\to\infty}\left(1 + \frac{1}{u}\right)^u}_{\text{This is }e}\right]^{\frac{1}{x}}$$Therefore: .$$f'(x) \;=\;\ln(e)^{\frac{1}{x}} \;=\;\frac{1}{x}\cdot\ln(e) \;=\;\frac{1}{x}$$

 

[Amer]

Thanks many :)