# Spivak on Exponential and Logarithmic Functions

I'm reading through parts of Spivak and was struggling with his discussion on exponential and logarithmic functions (Pg 340, 2008). In this portion of text he's trying to find the derivative of some sort of function that would behave as follows, $$f(x+y)=f(x)\cdot f(y)$$

So he assumes such a function exists and begins to take its derivative from first principles.

Below is an image of his text. I have a few questions.

What does he mean when he says, "The answer thus depends on...", why does it depend on $$f'(0)$$? And what happened to $$f(x)$$? Is he just saying this because f(x) is constant with respect to the actual limit?

Second, what did he do to start taking the derivative of the logarithm function? I'm probably missing something obvious but where did he get the expression, $$\frac{1}{f'(f^{-1}(x))}$$? Spivak shows that \begin{align} f'(x) &= f(x) \cdot \lim_{h \to 0}\dfrac{f(h) - 1}{h} \\ &= f(x) \cdot f'(0) \tag{*} \end{align} So, definitely, $$f'(x)$$ depends on $$f'(0)$$, simply because it appears on the RHS of the above formula. The reason he was able to "factor $$f(x)$$" out of the limit is because the limit is taken as $$h \to 0$$; but $$f(x)$$ doesn't depend on $$h$$, so it is a constant, and hence can be pulled out (this should be one of the first theorems of Chapter 5).

Note that he defined the logarithm as the inverse of $$f$$, so $$\log_{10} = f^{-1}$$. Hence, \begin{align} \log_{10}'(x) &= (f^{-1})'(x) \\ &= \dfrac{1}{f'(f^{-1}(x))} \tag{inverse function theorem} \\ &= \dfrac{1}{f(f^{-1}(x)) \cdot f'(0)} \tag{by *} \\ &= \dfrac{1}{x \cdot f'(0)} \end{align} If the inverse function theorem is still confusing, I suggest you review Chapter 12 of the text; if memory serves me right, that formula is part of Theorem 5 I believe.

• That clarifies things, I've been jumping around since I've only been reading since I started taking integral calculus. I'll take a look at the inverse function theorem. Apr 14, 2020 at 8:26

What he means by that is that what you get for the derivative $$f'(x)$$ of such a function depends on the limiting value of the quantity $$\frac{f(h)-1}{h}$$ as $$h$$ becomes infinitesimal, which is the derivative $$f'(0)$$ of the function at $$x=0$$ if you rewrite it thus $$\frac{f(h)-f(0)}{h-0},$$ where you can see that $$f(0)=1$$ from the functional equation $$f(x+y)=f(x)f(y)$$ by setting $$x=y=0.$$ You get two possibilities for $$f(0),$$ namely $$0$$ or $$1,$$ but we may rule out the first since it leads to the constant function $$0,$$ since we would then have, by setting $$y=0$$ in the functional equation, $$f(x)=f(x)f(0)=0.$$ Thus if we want an interesting function, then we should take $$f(0)=1.$$

The quantity $$f'(x)$$ depends on $$f'(0)$$ because we need to evaluate the limit $$f'(0)$$ (if it exists) to be able to define $$f'(x)$$ as $$f(x)f'(0).$$

Nothing happened to $$f(x).$$ He's only concentrating on what we need to find, namely the limit $$f'(0).$$

Yes, you can see that $$f(x)$$ does not depend on $$h.$$ That's why it's not involved with the quantity $$f'(0)$$ and he was able to factor it out.

To see how he differentiated the inverse function, set $$y=f^{-1}(x).$$ Then we have, by definition (or by applying $$f$$ to both sides), that $$f(y)=x.$$ Now, differentiating, and keeping in mind that $$y$$ is dependent on $$x,$$ gives $$f'(y)y'=1,$$ so that $$y'=\frac{1}{f'(y)}=\frac{1}{f'(f^{-1}(x))}.$$

Hope this helps.