The Ripple That Spreads
A stone drops into a calm pond. The outermost ripple's radius grows at a steady rate, $r(t)=3t$ metres after $t$ seconds, and the enclosed area is $A=\pi r^{2}$. How fast is the area growing at $t=2$ seconds? The area depends on the radius, which depends on time — we must chain the rates together.
radius 6.0 m · $\dfrac{\text{d}r}{\text{d}t}=3$ · $\dfrac{\text{d}A}{\text{d}r}=2\pi r$ = 37.7 · $\dfrac{\text{d}A}{\text{d}t}=6\pi r$ = 113.1 m²/s
At $t=2$: $r=6$, so $\dfrac{\text{d}A}{\text{d}t}=6\pi\cdot 6=36\pi\approx 113.1$ m²/s.
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. $\dfrac{d}{dx}\big(x^5\big)$ = ?
R2. $\dfrac{d}{dx}\big(e^{x}\big)$ = ?
R3. The inside function $u$ of $y=(3x+1)^2$ is:
The area $A$ changes as the radius $r$ changes, and the radius changes as time changes. Each rate multiplies the next:
Writing the inside function as $u$ and the outside as $y(u)$ gives the rule in its compact form:
Differentiate the outer function, then multiply by the derivative of the inner function.
With $u$ a function of $x$ and $u'=\dfrac{du}{dx}$:
| Function | Derivative |
|---|---|
| $e^{u}$ | $u'\,e^{u}$ |
| $\ln u$ | $\dfrac{u'}{u}$ |
| $\sin u$ | $u'\cos u$ |
| $\cos u$ | $-u'\sin u$ |
| $\tan u$ | $u'\sec^{2}u$ |
| $u^{n}$ | $n u^{n-1}u'$ |
Each row is the basic derivative from Lesson 1 with an extra factor of the inner derivative $u'$.
$u=2x,\ \dfrac{dy}{du}=e^{u}$ → $\dfrac{dy}{dx}=2\cdot e^{2x}=2e^{2x}$.
$u=2x,\ \dfrac{dy}{du}=\cos u$ → $\dfrac{dy}{dx}=2\cos 2x$.
$u=2x+1,\ \dfrac{dy}{du}=5u^{4}$ → $\dfrac{dy}{dx}=5(2x+1)^{4}\cdot 2=10(2x+1)^{4}$.
Always ask: what is the inside $u$, and what is its derivative $u'$?
Groups of 4: tap a claimed derivative to pick it up, then tap a zone to file it — "✓ Chain rule right" or "✗ Chain-rule slip". Discuss why each card holds or misses the inner factor, then press Check cards. Six claims across e, ln, trig and power composites.
Each machine shows a composite y = f(g(x)). Pull two levers — tap the correct OUTER derivative f′(g(x)) and the correct INNER derivative u′ — then Run the machine. It feeds both through to assemble dy/dx. Four machines; score appears on the timer.
Differentiate $y=\ln(x^2+1)$.
$u=x^2+1$, so $u'=2x$.
$\dfrac{dy}{dx}=\dfrac{2x}{x^2+1}$.
Differentiate $y=\tan(x^2+1)$.
$u=x^2+1$, so $u'=2x$.
$\dfrac{dy}{dx}=2x\sec^{2}(x^2+1)$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. $\dfrac{d}{dx}\big(e^{3x}\big)$ = ?
Q2. $\dfrac{d}{dx}\big(\sin 3x\big)$ = ?
Q3. $\dfrac{d}{dx}\big((2x+1)^3\big)$ = ?
Q4. $\dfrac{d}{dx}\big(\ln(2x+1)\big)$ = ?
Work each on paper, then reveal the solution to check your working.
Differentiate $y=e^{2x}$.
Reveal Solution
$u=2x$, $u'=2$, so $\dfrac{dy}{dx}=2e^{2x}$.
Differentiate $y=\sin 2x$.
Reveal Solution
$\dfrac{dy}{dx}=2\cos 2x$.
Differentiate $y=e^{x^2}$.
Reveal Solution
$u=x^2$, $u'=2x$, so $\dfrac{dy}{dx}=2x\,e^{x^2}$.
Differentiate $y=\sqrt{\cos x}$.
Reveal Solution
$y=(\cos x)^{1/2}$, so $\dfrac{dy}{dx}=\tfrac12(\cos x)^{-1/2}\cdot(-\sin x) =-\dfrac{\sin x}{2\sqrt{\cos x}}$.
Differentiate $y=e^{\cos 2x}$.
Reveal Solution
Double chain: $\dfrac{d}{dx}(e^{\cos 2x})=e^{\cos 2x}\cdot(-\sin 2x)\cdot 2 =-2\sin 2x\,e^{\cos 2x}$.
Exit Ticket
Check the learning objectives you feel confident about: