If we examine the inverse function $f^{-1}=log_{10}$, the whole situation appears in a new light:
$$\begin{align} log_{10}'(x)&=\frac1{f'\left(f^{-1}(x)\right)}\\ &=\frac1{\alpha\cdot f\left(f^{-1}(x)\right)}=\frac1{\alpha x} \end{align}$$
The derivative of $f^{-1}$ is about as simple as one could ask! And, what is even more interesting, of all the integrals $\displaystyle\int_a^bx^n\,\mathrm dx$ examined previously, the integral $\displaystyle\int_a^bx^{-1}\,\mathrm dx$ is the only one which we cannot evaluate. Since $\log_{10}1=0$ we should have $$\frac1\alpha\int_1^x\frac1t\,\mathrm dt=\log_{10}x-\log_{10}1=\log_{10}x$$
$\alpha$ is defined as follows:
…then ($*$) will imply that $f(x)=10^x$ for rational $x$, and $10^x$ could be defined as $f(x)$ for other $x$; in general $f(x)$ will equal $[f(1)]^x$ for rational $x$.
One way to find such a function is suggested if we try to solve an apparently more difficult problem: find a differentiable function $f$ such that
$$f(x+y)=f(x)\cdot f(y)\quad\text{for all }x\text{ and }y\text,\\f(1)=10$$
Assuming that such a function exists, we can try to find $f'$—knowing the derivative of $f$ might provide a clue to the definition of $f$ itself. Now
$$\begin{align} f'(x)&=\lim_{h\to0}\frac{f(x+h)-f(x)}h\\ &=\lim_{h\to0}\frac{f(x)\cdot f(h)-f(x)}h\\ &=f(x)\cdot\lim_{h\to0}\frac{f(h)-1}h\\ \end{align}$$
The answer thus depends on $$f'(0)=\lim_{h\to0}\frac{f(h)-1}h\text;$$ for the moment assume this limit exists, and denote it by $\alpha$. Then $$f'(x)=\alpha\cdot f(x)\quad\text{for all }x$$
Even if $\alpha$ could be computed, this approach seems self-defeating. The derivative of $f$ has been expressed in terms of $f$ again.
Why is it here? I can't make any sense out of its use.
