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Derivative of ln(x)

Amer

Active member
Mar 1, 2012
275
Is it possible to find the derivative of ln(x) by the definition how ?

[tex]f'(x) = \lim _ {h\rightarrow 0 } \frac{f(h+x) - f(x)}{h} [/tex]
 

Reckoner

Member
Jun 16, 2012
45

Siron

Active member
Jan 28, 2012
150
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

Well-known member
Feb 2, 2012
409
Hello, Amer!

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

[tex]f'(x)\:=\:\lim_{h\to0}\frac{f(x+h)- f(x)}{h}[/tex]

[tex]f(x+h) - f(x) \;=\; \ln(x+h) - \ln(x) \;=\;\ln\left(\frac{x+h}{x}\right) \;=\;\ln\left(1 + \frac{h}{x}\right) [/tex]


[tex]\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}}[/tex]

. . . . . . . . . . . . [tex]=\;\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}} [/tex]


[tex]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}} [/tex]

Let [tex]u \,=\,\frac{x}{h}[/tex] . . Note: if [tex]h\to0[/tex], then [tex]u \to\infty[/tex]
We have: .[tex]\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}} [/tex]


Therefore: .[tex]f'(x) \;=\;\ln(e)^{\frac{1}{x}} \;=\;\frac{1}{x}\cdot\ln(e) \;=\;\frac{1}{x}[/tex]
 

Amer

Active member
Mar 1, 2012
275
Thanks many :)