From Slope to the Limit of a Rate
A slope is a rate of change between two points. Slide point B towards point A and the slope of the line joining them chases a single stable value — the instantaneous rate. That limiting value is the derivative:
This definition is the engine behind every result in today's lesson — you will see it at work in each Key Concepts derivation below.
secant slope 5.93 → tangent slope $e^{1}$ ≈ 2.72
For any base $a>0$, the slope of $y=a^{x}$ at $x=0$ is a fixed number depending only on $a$: $L(a)=\displaystyle\lim_{h\to0}\dfrac{a^{h}-1}{h}$. Slide the base and watch that slope at $x=0$. Exactly one base makes it equal to 1.
slope at $x=0$: 0.693 < 1 — too shallow
A deposit of $100$ earns $100\%$ per year, compounded $n$ times a year, so after one year it is worth $100\big(1+\tfrac{1}{n}\big)^{n}$:
| payments per year $n$ | multiplier $\big(1+\tfrac{1}{n}\big)^{n}$ | value after 1 year |
|---|---|---|
| 1 | 2.000000 | 200.00 |
| 4 | 2.441406 | 244.14 |
| 12 (monthly) | 2.613035 | 261.30 |
| 365 (daily) | 2.714567 | 271.46 |
| compounding continuously ($n\to\infty$) | $e\approx2.7182818$ — the ceiling | 271.83 |
More payments, more money — yet the multiplier climbs towards a ceiling it can never pass. Even compounding every second ($n=31\,536\,000$) reaches only $2.7182818$, a whisker below the ceiling. Bernoulli had discovered a new number, $e=\displaystyle\lim_{n\to\infty}\big(1+\tfrac{1}{n}\big)^{n}\approx2.7182818$ — and no derivatives were involved. This is where $e$ comes from.
Differentiating any exponential from first principles — the very number the slider above prints:
Leibniz could compute this constant — $L(2)\approx0.6931$, $L(10)\approx2.3026$ — but for years nobody could name it. The hunt was on for the one base with $L(a)=1$: an exponential that is exactly its own derivative.
Test the value $L(2)=0.6931$: since $e^{0.6931}\approx2.000$, it answers “$e$ to what power gives $2$?” — that is the logarithm of $2$ to base $e$. Euler proved this in general, $L(a)=\ln a$, so the hunted base is the solution of $\ln a=1$, namely $a=e$ — Bernoulli's number.
And $e'(0)=\ln e=1$ is no special favour to $e$: it is the same fact as $\log_{2}2=1$ or $\log_{10}10=1$ — every logarithm equals 1 on its own base.
Two independent discoveries — a ceiling from banking, a constant from calculus — met on the same number. Feed $e'(0)=1$ into $\dfrac{d}{dx}e^{x}=e^{x}\cdot e'(0)$ and you get $\dfrac{d}{dx}e^{x}=e^{x}$. It is also why $\ln$ is called the natural logarithm — it is the constant that calculus itself produces.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. $\dfrac{d}{dx}\big(x^3\big)$ = ?
R2. $\dfrac{d}{dx}(5)$ = ?
R3. $\dfrac{d}{dx}\big(x^5 + 2x\big)$ = ?
R4. $\dfrac{d}{dx}\Big(\dfrac{1}{x}\Big)$ = ?
R5. $\dfrac{d}{dx}\big((2x+1)^5\big)$ = ?
| Function $f(x)$ | Derivative $f'(x)$ |
|---|---|
| $e^x$ | $e^x$ |
| $\ln x$ | $\dfrac{1}{x}$ |
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\tan^{-1}x$ | $\dfrac{1}{1+x^2}$ |
You will not just memorise this — below you derive all six: the two whose slope echoes their height, the log and the trig pair by inversion and ratio. Understanding makes them impossible to forget.
Together (group): drag the point along $y=e^x$. Read the slope of the tangent that the graph prints. Compare it with the height $e^x$. What do you notice?
$\displaystyle \dfrac{d}{dx}e^x = \lim_{h\to 0}\dfrac{e^{x+h}-e^x}{h} = e^x \lim_{h\to 0}\dfrac{e^{h}-1}{h} = e^x \lim_{h\to 0}\dfrac{e^{0+h}-e^{0}}{h} = e^x \, e'(0)$ (since $1=e^0$, so $e^{h}-1 = e^{0+h}-e^{0}$).
$e$ is defined so that $e'(0)=1$; hence $\dfrac{d}{dx}e^x = e^x \cdot 1 = e^x$.
$e^x$ is the only function that is its own rate of change.
Partner check: $\ln x$ is the inverse of $e^x$. Drag the point: the tangent slope looks small for large x — the graph flattens as $x$ grows. Guess the slope at $x=2$ before you read the printed value.
Write $y=\ln x$. Then $e^y = x$.
$\dfrac{d}{dx}\big(e^y\big)=\dfrac{d}{dx}(x) \;\Rightarrow\; e^y\dfrac{dy}{dx}=1$.
$\dfrac{dy}{dx}=\dfrac{1}{e^y}=\dfrac{1}{x}$, for $x>0$.
Whole-class, then pairs: $\sin x$ starts flat at 0, rises fastest at $x=\dfrac{\pi}{2}$, then flattens again — the pattern of a $90^\circ$-shifted sine. Drag $x$: the tangent slope of $\sin x$ at every point equals $\cos x$ there.
$\displaystyle \dfrac{d}{dx}\sin x = \lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h} = \sin x\lim_{h\to0}\dfrac{\cos h -1}{h} + \cos x\lim_{h\to0}\dfrac{\sin h}{h}$
Geometrically (area comparison in the unit circle): $\displaystyle \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1$ and $\displaystyle \lim_{\theta\to0}\dfrac{1-\cos\theta}{\theta}=0$. Both curves are undefined at $\theta=0$, but each settles on a value (the hollow dot) as $\theta$ nears 0 from either side — that value is the limit:
$\dfrac{d}{dx}\sin x = \sin x(0) + \cos x(1) = \cos x$, and similarly $\dfrac{d}{dx}\cos x = -\sin x$.
In pairs: $\tan x$ is the ratio of the pair you already know ($\tan x = \dfrac{\sin x}{\cos x}$), and $\tan^{-1}x$ is the inverse of $\tan x$, just as $\ln x$ inverts $e^{x}$. Their derivatives are given in the table — here is why the table can trust them.
$\tan x = \dfrac{\sin x}{\cos x}$ — the ratio of two derivatives you already hold. When you divide like this, the $\sin x$ and $\cos x$ pieces cancel into the clean result given in the table: $\dfrac{d}{dx}\tan x = \sec^{2}x$.
The careful proof of that cancellation is the job of the quotient rule, which we derive in Lesson 4. Until then, take $\sec^{2}x$ as given — and use it.
Exactly as $\ln x$ undoes $e^{x}$ to give $\tfrac{1}{x}$, the inverse of $\tan$ has its own simple ratio. The table states it as given: $\dfrac{d}{dx}\big(\tan^{-1}x\big) = \dfrac{1}{1+x^{2}}$.
A clue to believe it: a small change in $x$ moves $\tan^{-1}x$ by about $\tfrac{1}{1+x^{2}}$ — tiny far from $0$ (the flat branches), largest at $0$. The strict derivation uses inverse / implicit differentiation, reached in Lesson 6.
You do not prove $\sec^{2}x$ or $\tfrac{1}{1+x^{2}}$ today. You recognise them in the table and apply them alongside $e^{x}$, $\ln x$, $\sin x$ and $\cos x$. Lessons 4 and 6 later repay the promise with formal proofs.
In teams of 4. Tap a function card on the left, then the matching derivative on the right — a coloured line links them and numbers a chip on each side. Get all 6 correct to unlock the summary. Wrong pairs turn red for you to fix.
Each attempt below contains one wrong step. Find it, tap it, then pick the correct derivative. Your partner checks your reasoning out loud after you answer. The clock starts on your first verdict.
Differentiate $y = e^x + \ln x + \sin x$.
The function is a sum of three independent terms: $e^x$, $\ln x$, $\sin x$. Each differentiates on its own.
$\dfrac{d}{dx}(e^x)=e^x$, $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$, $\dfrac{d}{dx}(\sin x)=\cos x$.
$\dfrac{dy}{dx} = e^x + \dfrac{1}{x} + \cos x$.
Differentiate $y = 4e^x - 2\ln x + 3\cos x$.
A linear combination: $4e^x$, $-2\ln x$, $3\cos x$.
$4\dfrac{d}{dx}(e^x)=4e^x$, $-2\dfrac{d}{dx}(\ln x)=-\dfrac{2}{x}$, $3\dfrac{d}{dx}(\cos x)=-3\sin x$.
$\dfrac{dy}{dx} = 4e^x - \dfrac{2}{x} - 3\sin x$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. $\dfrac{d}{dx}(3e^x-5)$ = ?
Q2. $\dfrac{d}{dx}(2\ln x+3)$ = ?
Q3. $\dfrac{d}{dx}\big(3\sin x+2\cos x-\tan x\big)$ = ?
Q4. $\dfrac{d}{dx}\big(5\tan^{-1}x-2\big)$ = ?
Work each on paper, then reveal the solution to check your working.
Differentiate $y = \ln\Big(\dfrac{1}{x}\Big)$.
Reveal Solution
$\ln\Big(\dfrac{1}{x}\Big)=\ln(x^{-1})=-\ln x$
$\dfrac{dy}{dx}=-\dfrac{1}{x}$
Differentiate $y = 3e^x$.
Reveal Solution
$\dfrac{d}{dx}(e^x)=e^x$, so the constant multiple scales it: $\dfrac{dy}{dx}=3\dfrac{d}{dx}(e^x)=3e^x$.
Differentiate $y = 5\ln x - 4\cos x + 2$.
Reveal Solution
$\dfrac{dy}{dx} = \dfrac{5}{x} + 4\sin x$ ($2$ vanishes as a constant).
Exit Ticket
Check the learning objectives you feel confident about: