Concentration Under Change
A chemist monitors a solution whose solute mass $m(t)=2t+1$ g and volume $V(t)=t+2$ L both grow with time. Concentration is a quotient: $c=\dfrac{m}{V}$. Both the top and the bottom are changing — how fast does the quotient change?
A quotient is a product with a negative power: $c=m\cdot V^{-1}$. Drag $t$ — the rate of the quotient combines the rates of both parts.
m = 3.0 g · V = 3.0 L · c = 1.00 g/L
$\dfrac{\text{d}c}{\text{d}t}=\dfrac{m'V-mV'}{V^{2}}$ = 0.33 g/L per 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. The product rule for $(uv)$ is:
R2. Rewrite $\dfrac{1}{x^{3}}$ using a negative index:
R3. $\dfrac{d}{dx}\big(x^{-2}\big)$ = ?
A fraction is really a product: $\dfrac{u}{v}=u\cdot v^{-1}$. So the product rule from Lesson 3 already knows how to differentiate quotients — we just tidy the result into one fraction.
Say it as: low d-high minus high d-low, over low squared. The order of the numerator matters — swap it and the sign flips.
| Function $y$ | Derivative $\dfrac{dy}{dx}$ |
|---|---|
| $\dfrac{x}{x-1}$ | $-\dfrac{1}{(x-1)^2}$ |
| $\dfrac{\sin x}{x}$ | $\dfrac{x\cos x-\sin x}{x^2}$ |
| $\dfrac{e^x}{\cos x}$ | $\dfrac{e^x(\cos x+\sin x)}{\cos^2 x}$ |
| $\dfrac{e^x}{x}$ | $\dfrac{e^x(x-1)}{x^2}$ |
Top and bottom keep their own chain rules inside the formula when they are composites.
$u=e^x,\ u'=e^x$; $v=x,\ v'=1$, so $\dfrac{dy}{dx}=\dfrac{e^x\cdot x-e^x\cdot 1}{x^2}=\dfrac{e^x(x-1)}{x^2}$.
Set $x-1=0$ (since $e^x\ne 0$): $x=1$, $y=e$. Stationary point $(1,e)$.
Only the numerator of the derivative can vanish — the denominator is always positive (or undefined).
Groups of 4: tap a quotient on the left, then its derivative on the right — a coloured line links each pair. Six quotients covering rational, e and trig ratios. Wrong pairs turn red to fix.
Each attempt below claims a derivative. Decide whether the quotient rule was applied correctly, then explain your verdict out loud to a partner.
The curve $y=\dfrac{x}{e^x}$ has one stationary point. Find its exact coordinates.
$\dfrac{dy}{dx}=\dfrac{1\cdot e^x-x\cdot e^x}{(e^x)^2}=\dfrac{e^x(1-x)}{e^{2x}}=\dfrac{1-x}{e^x}$.
$1-x=0\ \Rightarrow\ x=1$, and $y=\dfrac{1}{e}$. Stationary point $\left(1,\dfrac{1}{e}\right)$.
Differentiate $y=\dfrac{2x}{3x-1}$.
$u=2x,\ u'=2$ and $v=3x-1,\ v'=3$.
$\dfrac{dy}{dx}=\dfrac{2(3x-1)-2x\cdot 3}{(3x-1)^2}=\dfrac{6x-2-6x}{(3x-1)^2}=-\dfrac{2}{(3x-1)^2}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. $\dfrac{d}{dx}\left(\dfrac{x}{x+1}\right)$ = ?
Q2. $\dfrac{d}{dx}\left(\dfrac{x^2}{x+1}\right)$ = ?
Q3. $\dfrac{d}{dx}\left(\dfrac{e^x}{x+1}\right)$ = ?
Q4. Find the gradient of $y=\dfrac{x}{x-1}$ at $(0,0)$.
Work each on paper, then reveal the solution to check your working.
Differentiate $y=\dfrac{x^3}{x^2+1}$.
Reveal Solution
$\dfrac{dy}{dx}=\dfrac{3x^2(x^2+1)-x^3\cdot 2x}{(x^2+1)^2}=\dfrac{x^4+3x^2}{(x^2+1)^2}=\dfrac{x^2(x^2+3)}{(x^2+1)^2}$.
Differentiate $y=\dfrac{x-3}{x-4}$.
Reveal Solution
$\dfrac{dy}{dx}=\dfrac{1(x-4)-(x-3)\cdot 1}{(x-4)^2}=-\dfrac{1}{(x-4)^2}$.
Differentiate $y=\dfrac{\sin x}{x}$.
Reveal Solution
$\dfrac{dy}{dx}=\dfrac{x\cos x-\sin x\cdot 1}{x^2}=\dfrac{x\cos x-\sin x}{x^2}$.
The curve $y=\dfrac{x^2}{2x+1}$ has stationary points. Find their coordinates and identify their nature.
Reveal Solution
$\dfrac{dy}{dx}=\dfrac{2x(2x+1)-x^2\cdot 2}{(2x+1)^2}=\dfrac{2x^2+2x}{(2x+1)^2}=\dfrac{2x(x+1)}{(2x+1)^2}$.
Stationary when $x=0$ or $x=-1$. Points $(0,0)$ and $(-1,-1)$. Sign of the numerator near each point shows $(0,0)$ is a minimum and $(-1,-1)$ is a maximum.
Exit Ticket
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