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I trying to find any good example of derivation of $\lg (x)$ from the very begining. I know that $(\lg (x))' = \frac{1}{x}$ but WHY? I cannot get the same answer by calculating it by my hands.

$$ \lim_ {h \to 0} \dfrac{\lg (x+h) - \lg (x)}{h} $$

How should i solve this limit?

Kemsikov
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  • Try to use sandwich theorem. – Anwesha1729 Dec 28 '20 at 17:14
  • This is often shown with the inverse function theorem: Since $\exp$ is differentiable its inverse function $\log$ is also differentiable and $\log'(x) = \frac{1}{\exp'(\log(x))} = \frac{1}{x}.$ This however doesn't answer the question how to see this using only the difference quotient alone. – Jannik Pitt Dec 28 '20 at 17:16
  • Are you asking for a logarithm with base $10$, or base $e$ ($\ln$)? Anyway, all logarithms are equal with multiplication by a constant (e.g. $\log_{10} (x) = \ln(x) \cdot \frac{1}{\ln 10}$... And then you can use the fact you've stated for the derivative of $\ln$ – talbi Dec 28 '20 at 17:16

3 Answers3

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Continuing from your work, we can use logarithm rules, and get

$$ \lim_{h \to 0} \frac{1}{h} \ln \left ({\frac{x+h}{x}} \right ) = \ln \left ( 1 + \frac{h}{x} \right )^\frac{1}{h} $$

We can do a switch of variables, e.g. $k = \frac{1}{h}$ and get

$$ \lim_{k \to \infty} \ln \left (1 + \frac{1}{kx} \right )^k $$

You can see that whats inside the $\ln$ approaches $e^{\frac{1}{x}}$. Therefore, $$ (\ln x)' = \ln e^{\frac{1}{x}} = \frac{1}{x}$$

Anyway, this is a "crude" way to prove the derivative of $\ln$. For the derivative of any logarithm with a different base, simply note that $$ \log_a (x) = \ln(x) \cdot \frac{1}{\ln a} $$

Finally, also check out this.

talbi
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You can grind through it like this:

$$\lim_{h \to 0} \frac{\ln(x+h)-\ln (x)}{h} = \lim_{h\to 0} \frac{\ln\left(\frac{x+h}{x}\right)}{h} $$

$$=\lim_{h\to 0} \frac{\ln\left(1+\frac{h}{x}\right)}{h} = \lim_{h\to 0} \ln\left(\exp\left( \frac{\ln\left(1+\frac{h}{x}\right)}{h}\right)\right)$$

$$= \lim_{h\to 0} \ln\left( \left(1+\frac{x}{h}\right)^{1/h}\right). $$

Now substitute $y =x/h$, so when $h\to 0$, then $y\to \infty.$ Also $h = x/y$. The above

$$ = \lim_{y\to \infty} \ln\left( \left(1+\frac{1}{y}\right)^{\frac{y}{x}}\right). $$

Since the (or one) definition of $e$ is $e=\lim_{y\to \infty} \left(1+\frac{1}{y}\right)^y,$ and since $\ln x$ is continuous, we have

$$= \ln\left( \left(\lim_{y\to \infty}\left(1+\frac{1}{y}\right)^y\right)^{1/x}\right)=\ln e^{1/x} = \frac{1}{x}. $$

B. Goddard
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$\log_{10}(x)=\frac1{\ln 10}\ln x$, therefore $\log_{10}'(x)=\frac1{x\ln 10}$ by what you've already established for $\ln$. If you haven't established what the derivative of $\ln$ is, apply identity $(f^{-1})'=\frac1{f'\circ f^{-1}}$ to $f(x)=\exp(x)$, obtaining $$\ln'(x)=\frac1{\exp'(\ln x)}=\frac1{\exp(\ln x)}=\frac1x.$$