Prove the derivative of the natural logarithm using the limit definition.

Question

I know how to prove that the derivative of $\ln(x)$ is ${1\over x}$ using the definition $f'(x) = {f(x+h) - f(x) \over h}$ but I have ran into trouble proving that the derivative of $\ln(f(x))$ is ${f'(x) \over f(x)}$. Is there a way to prove this strictly using the definition of the derivative?

Answer 1

Basically what you have proved is that $\ln(x+h) = \ln(x) + h/x + h\phi(h)$ where $\lim_{h\to0}\phi(h) = 0$ (the proof of that is dependent on the actual definition for $\ln x$, but would be trivial if we define it as $\ln x = \int_1^x dt/t$). You also know by assumtion that $f(x+h) = f(x) + f'(x)h + h\psi(h)$ where $\lim_{h\to0}\psi(h) = 0$. Now just put that in:

$\ln(f(x+h)) = \ln(f(x) + f'(x)h+h\psi(h)) = \ln(f(x))+{f'(x)h+h\psi(h)\over f(x)} + h\phi((hf'(x) + h\psi(h)))$

Now plug that into the difference formula:

${\ln(f(x+h)) - \ln(f(x))\over h} = {f'(x)\over f(x)} + {\psi(h)\over f(x)} + \phi(hf'(x) + h\psi(h))$

Now as $h\to0$ we have that $hf'(x) + h\psi(h)\to0$ so it's clear that the limit of that expression is $f'(x)/f(x)$ as $h\to0$

Note the initial expressions that I claimed you have proven and assumed is due to:

$\psi(h) = {f(x+h) - f(x)\over h} - f'(x)$

Answer 2

First prove the chain rule $(g(f(x))'=f'(x)g'(f(x))$.

Indeed for any $h>0$,

$$g(f(x+h))=g(f(x)+h\frac{f(x+h)-f(x)}h)=g(f(x)+h'(x,h)),$$

then

$$\frac{g(f(x+h))-g(f(x))}h=\frac{g(f(x)+h'(x,h))-g(f(x))}{h'(x,h)}\frac{f(x+h)-f(x)}h$$

and taking the limit you get the requested formula (provided $f'(x)$ is defined, so that $h'(x,h)$ goes to zero at the same time as $h$).

With $g(x)=\ln(x)$, this gives

$$\ln(f(x))'=\frac{f'(x)}{f(x)}.$$

Answer 3

Consider that $$ \ln(f(x)) = y $$ so that $$ \mathrm{e}^y = f(x). $$ Then $$ (\mathrm{e}^y)' = \mathrm{e}^y \dfrac{\mathrm{d}y}{\mathrm{d}x} \quad\Longrightarrow\quad \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(\mathrm{e}^y)'}{\mathrm{e}^y} = \dfrac{f'(x)}{f(x)}. $$