Calculus I · The Function That Is Its Own Slope

Every exponential is already proportional to itself

Start with a plain exponential function, f(x)=axf(x) = a^x, where aa is some fixed positive number — maybe 22, maybe 1010. You met these in Book 1. Let us just take its derivative from scratch, the way we have taken every derivative so far: write down the difference quotient and see what happens.

f(x+h)−f(x)h=ax+h−axh.\frac{f(x+h) - f(x)}{h} = \frac{a^{x+h} - a^{x}}{h}.

Now use the one exponent rule that does all the work here: ax+h=ax⋅aha^{x+h} = a^x \cdot a^h. Factor the axa^x out:

axah−axh=ax⋅ah−1h.\frac{a^{x}a^{h} - a^{x}}{h} = a^{x}\cdot\frac{a^{h} - 1}{h}.

Look hard at what just happened, because it is the heart of the chapter. The axa^x came out front and did not change as hh shrank. The only part that depends on hh is the fraction ah−1h\dfrac{a^h - 1}{h}. So when we take the limit as h→0h \to 0:

ddxax=ax⋅(limh→0ah−1h)⏟a number that depends only on a.\frac{d}{dx}a^{x} = a^{x}\cdot \underbrace{\left(\lim_{h\to 0}\frac{a^{h}-1}{h}\right)}_{\text{a number that depends only on }a}.

That underbraced limit is just a number. It does not depend on xx at all. Let us give it a name and call it M(a)M(a), for the moment, because it is the mystery number attached to the base aa:

ddxax=M(a)⋅axwhereM(a)=limh→0ah−1h.\boxed{\;\frac{d}{dx}a^{x} = M(a)\cdot a^{x}\;}\qquad\text{where}\qquad M(a) = \lim_{h\to 0}\frac{a^{h}-1}{h}.

Read that boxed line out loud, because it is astonishing and most people never notice it. Every exponential function is already proportional to its own slope. The derivative of axa^x is just axa^x scaled by some fixed number M(a)M(a). We have almost answered our question for free. The only thing standing between us and a function that equals its own derivative is that pesky constant M(a)M(a) out front. If only we could find a base aa that made M(a)M(a) exactly equal to 11.

One more thing to notice before we hunt for that base. What is M(a)M(a), geometrically? Plug x=0x = 0 into the boxed rule. Since a0=1a^0 = 1, the slope of axa^x at x=0x = 0 is exactly M(a)⋅1=M(a)M(a)\cdot 1 = M(a). So:

M(a)M(a) is simply the slope of the curve y=axy = a^x as it crosses the yy-axis.

That gives us something we can actually look at. (Figure 5.1: the curves y=2xy=2^x, y=3xy=3^x, and a couple of others, each with its tangent line drawn where it crosses the yy-axis. The steeper the curve leaves the axis, the bigger M(a)M(a).)

Figure 5.1
Figure 5.1

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