# Derivative of a logarithm from first principles

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?

• 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 $\log$. – Patrick Stevens Aug 21 '15 at 11:16
• 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. – Henning Makholm Aug 21 '15 at 11:20
• @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... – 5xum Aug 21 '15 at 11:23
• @5xum, that also depends on the definition of $\exp$. In my course, it was defined as a power series. – Patrick Stevens Aug 21 '15 at 11:29

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

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

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

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

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

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