Introduction

In the previous post we built up why differentiating sine gives cosine, from a single rotation. Today we move to the other two pillars of differentiation: the exponential and the logarithm.

The story starts from a fact we met earlier. The exponential \(e^x\) with base \(e\) is its own derivative.

$$\frac{d}{dx}e^x = e^x$$

This is why the number \(e\) became the protagonist of calculus. But two natural questions remain.

  • What happens if we differentiate \(2^x\) or \(10^x\), whose base is not \(e\)? Why are they not clean?
  • Why does differentiating the logarithm \(\ln x\) make the odd-looking \(\dfrac{1}{x}\) pop out?

Remarkably, both answers come from one and the same root. Let us write the conclusions down first.

$$\frac{d}{dx}a^x = a^x\ln a, \qquad \frac{d}{dx}\ln x = \frac{1}{x}$$

The exponential \(a^x\) picks up a log factor \(\ln a\), and \(\ln x\) picks up a reciprocal \(\dfrac1x\). We will see why each takes that shape — by looking, not memorizing.

Here \(\ln\) is the logarithm with base \(e\), the natural logarithm. \(\ln a\) means "to what power must you raise \(e\) to get \(a\)." This meaning becomes the key part later.

1. Recap: what it means for the exponential with base e to be 'clean'

Notation first. \(e\) is a specific number, \(2.718\dots\) (the natural constant), and \(e^x\) is the exponential with that number as its base. "Differentiates cleanly" means its shape does not change under differentiation — the derivative equals the original function.

  • Intuition: an exponential grows faster the larger its value (just as a larger principal earns more interest). So differentiating any exponential returns 'the original function × some steady factor.'
  • Explanation: that factor differs by base, and the one base where it is exactly \(1\) is \(e\).
  • Formula: hence \(\dfrac{d}{dx}e^x = e^x\) — the factor is 1, so it is its own derivative.

The identity of this "factor" is the first key today. What is the factor for \(2^x\), and why is only \(e^x\)'s factor equal to 1?

2. Why other bases pick up a log factor — rewriting everything with base e

The key idea is a single one: any exponential, whatever its base, can be rewritten as an exponential with base \(e\).

By the definition of the natural logarithm, \(a = e^{\ln a}\) (since \(\ln a\) is "the power to which \(e\) must be raised to get \(a\)," raising \(e\) to it gives \(a\) back). Raise both sides to the power \(x\):

$$a^x = \left(e^{\ln a}\right)^x = e^{(\ln a)\,x}$$

Now \(a^x\) has turned into \(e^{(\ln a)x}\), an exponential with base \(e\). It is a layered structure, with the function \((\ln a)x\) sitting in the exponent slot in place of \(x\). A function inside a function is differentiated by the chain rule — differentiate the outer function, then multiply by the derivative of the inner one.

  • Intuition: the outer part \(e^{(\cdots)}\) is unchanged by differentiation. The inner part \((\ln a)x\) has a steady slope of \(\ln a\) in \(x\). So that \(\ln a\) pops out as the factor.
  • Explanation: outer derivative \(= e^{(\ln a)x} = a^x\), inner derivative \(= \ln a\). Multiply the two.
  • Formula:
$$\frac{d}{dx}a^x = e^{(\ln a)x}\cdot \ln a = a^x\ln a$$

That settles everything. The factor was \(\ln a\). Let us check base by base.

  • If \(a=e\), then \(\ln e = 1\), so the factor is 1, giving \(\dfrac{d}{dx}e^x = e^x\). That single line is why \(e^x\) is clean.
  • If \(a=2\), then \(\ln 2 \approx 0.693\), so \(\dfrac{d}{dx}2^x = 2^x\cdot 0.693\). It grows slightly less than the original (because \(2 < e\)).
  • If \(a=10\), then \(\ln 10 \approx 2.303\), so \(\dfrac{d}{dx}10^x = 10^x\cdot 2.303\). It is more than twice as steep as the original.

Below \(e\) the factor is under 1; above \(e\) it is over 1. So \(e\) is the special base on that boundary, the one whose factor is exactly 1.

3. Why differentiating the log gives a reciprocal — the speed at which area fills

Now for the logarithm. Differentiate \(\ln x\) and you get \(\dfrac1x\). Why does a fraction, which seems to have nothing to do with exponentials and logs, pop out?

The brightest picture is to see the logarithm as an area. The natural logarithm has this face:

$$\ln x = \int_1^x \frac{1}{t}\,dt$$

That is, \(\ln x\) is the area accumulated under the curve \(\dfrac1t\), from \(t=1\) to \(t=x\). (At \(x=1\) the area is 0, which matches \(\ln 1 = 0\).) And from the fact that integration and differentiation are opposites, the speed at which the accumulated area grows = the height of the curve right there.

  • Intuition: widen the area a tiny bit to the right, and the height of the thin strip newly added is exactly the speed at which the area is swelling at that instant.
  • Explanation: at \(x\), the height of the curve \(\dfrac1t\) is \(\dfrac1x\). So the speed at which the area \(\ln x\) grows — its slope — is \(\dfrac1x\).
  • Formula:
$$\frac{d}{dx}\ln x = \frac{1}{x}$$

Numbers make it real. At \(x=2\) the slope of the log is \(\dfrac12 = 0.5\), and the area accumulated by then is \(\ln 2\approx 0.693\). At \(x=4\) the slope flattens to \(\dfrac14=0.25\) and the area grows to \(\ln 4\approx 1.386\). As \(x\) grows, \(\dfrac1x\) shrinks, which is exactly where the famous gentleness of the logarithm — growing ever more slowly — comes from.

There is another route too. Since \(\ln x\) is the inverse function of \(e^x\), if \(y=\ln x\) then \(x=e^y\). Then \(\dfrac{dx}{dy}=e^y=x\), and flipping it gives \(\dfrac{dy}{dx}=\dfrac1x\). The fact that \(e^x\) is its own derivative is, here too, the key.

4. Check it yourself in the interactive below

Handle two scenes. One is that the factor of \(a^x\) is \(\ln a\); the other is that the slope of \(\ln x\) is the fill-speed of the area, \(\dfrac1x\).

  • [Exponential: factor = ln a] mode: change the base \(a\) with the slider, and the \(a^x\) curve and the tangent at one point change with it. Tangent slope ÷ value is the factor, and you can confirm it always equals \(\ln a\) exactly. Bring the base near \(e\approx2.72\) and the factor becomes 1, so slope = value — the moment the function is its own derivative.
  • [Log: slope = 1/x] mode: the area under the curve \(\dfrac1t\) from \(t=1\) is exactly \(\ln x\). Move \(x\) and both the accumulated area (\(=\ln x\)) and the current height \(\dfrac1x\) are shown. See with your eyes that this height is the speed at which the area swells — that is, the slope of \(\ln x\).
The exponential's factor is the natural log, the logarithm's slope is a reciprocal
In the first mode, changing the base keeps the ratio of tangent slope to value always equal to the natural-log value. In the second mode, moving x keeps the area under 1/t equal to ln x, whose height there is the slope of the log.

Handling it, you feel with your fingertips that neither the \(\ln a\) attached to \(a^x\) nor the \(\dfrac1x\) attached to \(\ln x\) is a forced rule — one is the inner slope that pops out when you rewrite the base as \(e\), the other is the speed at which an area fills.

5. Summary — both pillars run through e

  • The factor of \(a^x\) is \(\ln a\): any exponential can be rewritten as \(a^x = e^{(\ln a)x}\), and the chain rule pulls out the inner slope \(\ln a\) as the factor. At \(a=e\), \(\ln e=1\), so only \(e^x\) differentiates into itself.
  • The slope of \(\ln x\) is \(\dfrac1x\): \(\ln x\) is the area accumulated under \(\dfrac1t\), so the speed at which the area grows (= the curve height there) is exactly \(\dfrac1x\). The larger \(x\) is, the gentler it gets.
  • One root: both facts branch off from the single property that '\(e^x\) has itself as its derivative.' The exponential's factor and the logarithm's reciprocal both run through \(e\).

Whether the base is 2 or 10, an exponential borrows \(e\)'s clothes for a moment to be differentiated, and the logarithm reveals its slope through a different face, area. The \(\ln a\) and the \(\dfrac1x\) that look unrelated on the surface are in fact two shadows sharing the single axis \(e^x\).

Closing

Getting here, we have gathered almost all the basic parts of differentiation — polynomials, sine and cosine, and today's exponential and logarithm. Yet one piece is still missing. How do we differentiate a structure with another function nested layer upon layer inside a function, like \(\sin(x^2)\) or \(e^{3x}\)? Today, too, we quietly borrowed that principle (the chain rule) when differentiating \(a^x = e^{(\ln a)x}\); next time we will build that layered structure itself into a single picture.