Visual walkthrough — Derivatives of ln x and logₐ(x)

Before we start, one word we lean on: a slope is "how steep a curve is" — how much the height changes when you step a tiny bit to the right. A slope of $2$ means "step right by $1$, climb by $2$." A slope of $\frac{1}{2}$ means "step right by $1$, climb by only a half." That is the whole game: we want the slope of the curve $y=\ln x$ at every point.

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Step 1 — What is $\ln x$, really? (a picture, not a formula)

WHAT. $\ln x$ is defined as the inverse of the function $e^x$. Saying "$y=\ln x$" is exactly the same statement as "$e^y=x$." They are two ways to write one fact.

WHY. We are not allowed to use any slope formula yet — we don't have one. So we must anchor $\ln$ to something we do trust: $e^x$, whose slope law we take from Derivative of e^x and a^x (it is its own slope). Every result below is squeezed out of that single relation $e^y=x$.

PICTURE. Look at the figure. The purple curve is $e^x$. The coral curve is $\ln x$. They are mirror images across the dashed grey line $y=x$ — that is what "inverse" looks like: swap the horizontal and vertical axes and one curve becomes the other.

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Step 2 — Reading slope off the mirror

WHAT. A mirror across $y=x$ swaps the roles of "horizontal step" and "vertical step". So it swaps rise and run. And slope is $\frac{\text{rise}}{\text{run}}$ — so a mirror flips the slope upside down (takes its reciprocal).

WHY. This is the geometric heart of the whole page. We know the slope of $e^x$; we want the slope of its mirror image $\ln x$. If mirroring flips slope, we can get one from the other with no algebra at all.

PICTURE. At the point where $e^x$ has a little right-triangle "step" of run $1$, rise $m$ (slope $m$), the mirrored point on $\ln x$ has the same triangle with the legs swapped: run $m$, rise $1$ — slope $\frac{1}{m}$. The red triangle and its reflected teal copy show this leg-swap.

We now turn this picture into an equation. That is Step 3.

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Step 3 — Differentiate both sides of $e^y=x$

WHAT. Take the relation $e^y=x$ and ask "how does each side change as $x$ nudges right?" We apply the same operation $\frac{d}{dx}$ (="rate of change with respect to $x$") to both sides.

WHY. An equation stays true if you do the same thing to both sides. The right side $x$ changes at rate $1$ (step right by a bit, $x$ grows by exactly that bit). The left side $e^y$ changes too — but $y$ itself is silently sliding as $x$ moves. This is the implicit idea.

$\underset{\text{left side}}{\underbrace{\frac{d}{dx}\left(e^y\right)}}=\underset{=1}{\underbrace{\frac{d}{dx}(x)}}$

Here $\frac{d}{dx}(x)=1$ because $x$'s rate of change with respect to itself is $1$.

PICTURE. The figure shows the two curves stacked as a chain: $x\to y\to e^y$. A tiny push $dx$ at the bottom travels up through $y$ and out through $e^y$. We must track how big the push is at each stage — that is what the Chain Rule does next.

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Step 4 — The chain rule unpacks the left side

WHAT. The left side $e^y$ is a function of $y$, and $y$ is a function of $x$. The chain rule says: total rate = (rate of $e^y$ per unit $y$) $\times$ (rate of $y$ per unit $x$).

$\frac{d}{dx}\left(e^y\right)=\underset{\begin{array}{c}\text{how fast }{e}^y\\ \text{grows per unit }y\end{array}}{\underbrace{e^y}}\cdot \underset{\begin{array}{c}\text{how fast }y\\ \text{grows per unit }x\end{array}}{\underbrace{\frac{dy}{dx}}}$

WHY. We only know one slope law for the exponential: $e^y$ grows at rate $e^y$ per unit of $y$. But $x$ is the variable being pushed, and $y$ only moves a fraction $\frac{dy}{dx}$ as fast. Multiply the two fractions of speed together — that is the chain rule. Without it we'd be measuring speed in the wrong units.

PICTURE. Two gears meshed: turning the $x$-gear turns the $y$-gear at ratio $\frac{dy}{dx}$, which turns the $e^y$-gear at ratio $e^y$. The overall gearing multiplies. The equation from Step 3 now reads:

$e^y\cdot \frac{dy}{dx}=1.$

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Step 5 — Solve, and put $x$ back in

WHAT. We have $e^y\cdot \frac{dy}{dx}=1$. Divide both sides by $e^y$:

$\frac{dy}{dx}=\frac{1}{e^y}.$

Now recall the relation from Step 1: $e^y=x$. Substitute:

$\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}.$

WHY. We want the answer expressed in terms of where we are, $x$ — not in terms of the height $y$. The relation $e^y=x$ is the bridge that swaps $e^y$ for the plain number $x$. This is the exact same reciprocal from the mirror in Step 2: $\frac{dy}{dx}=\frac{1}{e^y}$ is "one over the slope of $e^x$", now written using $x$.

PICTURE. The tangent line to $\ln x$ drawn at three spots — $x=\frac{1}{2}$, $x=1$, $x=4$ — with its slope printed. Slopes are $2$, $1$, $\frac{1}{4}$: exactly $\frac{1}{x}$. Watch the tangent flatten as $x$ grows.

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Step 6 — The same answer from first principles (a sanity check)

WHAT. Slope is officially defined as a limit: shrink a step $h\to 0$ and watch the average slope settle.

$\frac{d}{dx}\ln x={\lim }_{h\to 0}\underset{\text{rise over run of a tiny chord}}{\underbrace{\frac{\ln (x+h)-\ln x}{h}}}$

Using the log law $\ln A-\ln B=\ln \frac{A}{B}$ from Logarithm Laws, and the substitution $t=h/x$:

$=\frac{1}{x}{\lim }_{t\to 0}\frac{\ln (1+t)}{t}=\frac{1}{x}\cdot 1=\frac{1}{x}.$

WHY. The pure geometric mirror argument (Steps 1–5) is beautiful but leans on $e^x$'s slope. This second route leans instead on the standard limit $\frac{\ln (1+t)}{t}\to 1$ — a completely independent foundation. When two independent roads reach $\frac{1}{x}$, we trust it.

PICTURE. The chord (a straight line joining two nearby points on $\ln x$) tilts and settles onto the tangent as $h$ shrinks. Its printed slope creeps toward $\frac{1}{x}$.

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Step 7 — Any base, and the empty basement of $\ln$

WHAT. For a general base $a$ (with $a>0,a\ne 1$), change base: ${\log }_{a}x=\frac{\ln x}{\ln a}$. Since $\ln a$ is just a fixed number, we divide the slope by it:

$\frac{d}{dx}{\log }_{a}x=\frac{1}{\ln a}\cdot \frac{1}{x}=\frac{1}{x\ln a}.$

WHY. We refuse to memorise a second rule. Every $\log$ is a rescaled copy of $\ln$: dividing the whole curve's height by the constant $\ln a$ divides every slope by $\ln a$ too. Setting $a=e$ gives $\ln e=1$, so the factor vanishes and we're back to $\frac{1}{x}$ — the "basement is empty."

PICTURE. Three logs — $\ln x$, ${\log }_{2}x$, ${\log }_{10}x$ — are the same shape squashed vertically by $\ln a$. At $x=1$ their slopes are $1$, $\frac{1}{\ln 2}\approx 1.44$, $\frac{1}{\ln 10}\approx 0.43$: a bigger base means a flatter, gentler curve.

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Step 8 — The degenerate & edge cases (never leave a gap)

WHAT & WHY. Three boundaries the smooth story quietly relies on:

• $x\to 0^{+}$ (left edge of the domain). Slope $\frac{1}{x}\to +\infty$: the curve dives down to $-\infty$ with a vertical tangent. It never reaches $x=0$; the $y$-axis is a wall (an asymptote). $\ln x$ simply does not exist for $x\le 0$ — matching $e^y=x$ never producing a non-positive $x$.
• $x=1$. $\frac{1}{1}=1$: the one point where the log-hill is as steep as a $45^{\circ }$ ramp. Here $\ln 1=0$, so the curve crosses the axis.
• Negative $x$ via $\ln |x|$. For $x<0$ we can still take $\ln |x|$. Its slope is still $\frac{1}{x}$ (now a negative number for $x<0$) — the formula extends cleanly to every $x\ne 0$, but only through the absolute value.

PICTURE. The full curve with the $x=0$ wall marked, the $45^{\circ }$ crossing at $x=1$ flagged, and the mirrored branch $\ln |x|$ for $x<0$ drawn faint.

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The one-picture summary

Everything at once: $e^x$ and $\ln x$ mirrored across $y=x$; a matched pair of step-triangles with legs swapped (slope $m$ becomes $\frac{1}{m}$); the tangent to $\ln x$ at a general $x$ labelled with slope $\frac{1}{x}$; and $e^y=x$ printed as the bridge.

Recall Feynman retelling — the whole walk in plain words

A logarithm is just a reflection of the exponential in a mirror tilted at $45^{\circ }$. A mirror swaps left–right steps with up–down steps, and slope is up-over-across — so a mirror turns every slope upside-down. The exponential's slope at a point of height $x$ is $x$ itself (it's the self-copying function). Flip that upside-down and you get $\frac{1}{x}$ — the slope of $\ln$. To be safe I did it a second, totally different way: I took tiny chords, used the log law to fold two logs into one, and rode the standard limit $\frac{\ln (1+t)}{t}\to 1$ to the finish — same $\frac{1}{x}$. Other bases are just the same hill squashed shorter by the fixed number $\ln a$, so every slope shrinks to $\frac{1}{x\ln a}$; base $e$ leaves the basement empty because $\ln e=1$. And the fine print: the hill only exists to the right of zero, at $x=1$ it's a perfect $45^{\circ }$ ramp, and it dives to a vertical wall as you approach $x=0$.

Recall Rapid self-check

Slope of $\ln x$ at $x=4$? ::: $\frac{1}{4}$. Slope of $\ln x$ at $x=1$? ::: $1$ (a $45^{\circ }$ crossing). Why does a mirror across $y=x$ invert slopes? ::: It swaps rise and run, and slope is rise/run. Slope of ${\log }_{2}x$ at $x=1$? ::: $\frac{1}{\ln 2}\approx 1.44$. Where does $\ln x$ have a vertical tangent? ::: As $x\to 0^{+}$, where $\frac{1}{x}\to \infty$.

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Connections

• Derivative of e^x and a^x — the self-copying slope we mirror.
• Implicit Differentiation — Steps 3–5.
• Chain Rule — the meshed gears of Step 4.
• Logarithm Laws — folds the chord in Step 6.
• Standard Limit (1+t)^{1/t} → e — the independent foundation of Step 6.
• Logarithmic Differentiation — where this slope law is put to work.