0
$\begingroup$

The usual example where learning about the derivative is obtaining it for $f(x)=x^2$ from first principles (see this for example).

I am stumped on how use first principles to obtain the derivative of a natural logarithm. We need:

$$\lim_{h\rightarrow0}\frac{\ln(x+h)-\ln x}{h}=\lim_{h\rightarrow0}\frac{\ln(1+\frac{h}{x})}{h}$$

Now I am stuck. Of course I know that Taylor expansion around very small $x$ of $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\ldots$, but that's not something one is supposed to know when learning first principles of differentiation. Is there something clever that I am missing?

$\endgroup$
  • 1
    $\begingroup$ The usual way, of course, is to take $\exp(\log(x))$ and differentiate it using the chain rule. I don't know of a more first-principlesy way. To use the Taylor series would be to assume the result, since it is built from the derivative of $\log$. $\endgroup$ – Patrick Stevens Aug 21 '15 at 11:16
  • 4
    $\begingroup$ It depends on what your definition of the logarithm is. In some developments you simply define $\log x = \int_1^x\frac1t\, dt$, in which case the derivative is directly by the Fundamental Theorem of Calculus. $\endgroup$ – Henning Makholm Aug 21 '15 at 11:20
  • 1
    $\begingroup$ @PatrickStevens But that's not from first principles. Besides, I learnt what the derivative of $e^x$ is through the chain rule and the derivative of $\ln x$, not the other way around... $\endgroup$ – 5xum Aug 21 '15 at 11:23
  • $\begingroup$ @5xum, that also depends on the definition of $\exp$. In my course, it was defined as a power series. $\endgroup$ – Patrick Stevens Aug 21 '15 at 11:29
6
$\begingroup$

You should know that $$\lim_{k \to 0} \frac{\log(1+k)}{k}=1$$ Then, calling $k= \frac hx$, you get $$\lim_{h \to 0} \frac{\log(1+\frac hx)}{h}= \lim_{h \to 0} \frac{\log(1+\frac hx)}{x\frac hx} = 1 \cdot \frac{1}{x}= \frac{1}{x}$$

$\endgroup$
  • $\begingroup$ I may be being obtuse, but how do we know that $\lim \frac{\log(1+k)}{k} = 1$ without knowing the derivative of $\log$? $\endgroup$ – Patrick Stevens Aug 21 '15 at 11:30
  • 2
    $\begingroup$ Take $\log$ to the fundamental limit $$\lim_{x \to 0} (1+x)^{\frac 1x} = e$$ which is a (almost) direct consequence of definition of $e$: $$e= \lim_{t \to \infty} \left( 1+\frac 1t \right)^t$$ $\endgroup$ – Crostul Aug 21 '15 at 11:34
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

If $w=\log x$ then $e^w = x$.

$$\frac {\Bbb d(\Bbb e^w)} {\Bbb dx} = 1 \implies \frac {\Bbb d(\Bbb e^w)} {dw} \frac {\Bbb dw} {\Bbb dx} = 1 \implies \Bbb e^w \frac {\Bbb dw} {\Bbb dx} = 1 \implies \frac {\Bbb dw} {\Bbb dx} = \frac 1 {\Bbb e^w} = \frac 1 x \implies \frac {\Bbb d(\log x)} {\Bbb dx} = \frac 1 x .$$

$\endgroup$
  • $\begingroup$ (The $\mathbb d$ and $\mathbb e$ are over the top.) $\endgroup$ – Simon S Aug 21 '15 at 12:37
  • $\begingroup$ Thanks - that's much nicer $\endgroup$ – logicmonkey Aug 21 '15 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.