Product & Quotient Rules
Differentiating products and fractions of functions · §14.1
Product Rule
\(\dfrac{d}{dx}[uv] = u\dfrac{dv}{dx}+v\dfrac{du}{dx}\)
Use when two functions are multiplied together. Identify u and v, differentiate each, apply the formula.
Quotient Rule
\(\dfrac{d}{dx}\left[\dfrac{u}{v}\right] = \dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^2}\)
"Low d-High minus High d-Low, all over Low squared." v is the denominator.
Differentiate \(y = x^3(2x+1)^4\) (product rule)
Set up
\(u=x^3,\;v=(2x+1)^4\)
\(u'=3x^2,\;v'=4(2x+1)^3\cdot2=8(2x+1)^3\)
\(u'=3x^2,\;v'=4(2x+1)^3\cdot2=8(2x+1)^3\)
Apply
\(\dfrac{dy}{dx}=x^3\cdot8(2x+1)^3+3x^2\cdot(2x+1)^4\)
Factorise
\(=x^2(2x+1)^3[8x+3(2x+1)]\)
✓
\(=x^2(2x+1)^3(14x+3)\)
Differentiate \(y = \dfrac{x^2+1}{3x-2}\) (quotient rule)
Set up
\(u=x^2+1,\;v=3x-2\)
\(u'=2x,\;v'=3\)
\(u'=2x,\;v'=3\)
Apply
\(\dfrac{dy}{dx}=\dfrac{(3x-2)(2x)-(x^2+1)(3)}{(3x-2)^2}\)
Expand
\(=\dfrac{6x^2-4x-3x^2-3}{(3x-2)^2}=\dfrac{3x^2-4x-3}{(3x-2)^2}\)
Derivatives of Trig, Exponential & Log
Standard derivatives · With chain rule · §14.2
Standard derivatives to memorise
\(\sin x\)\(\cos x\)
\(\cos x\)\(-\sin x\)
\(\tan x\)\(\sec^2 x\)
\(e^x\)\(e^x\)
\(e^{ax}\)\(ae^{ax}\)
\(\ln x\)\(\dfrac{1}{x}\)
With chain rule: \(\dfrac{d}{dx}\sin(f(x))=\cos(f(x))\cdot f'(x)\). Same pattern applies to cos, tan, e^x, ln.
Differentiate \(y = e^{3x^2}\)
Chain rule
\(\dfrac{dy}{dx}=e^{3x^2}\cdot 6x\)
✓
\(=6xe^{3x^2}\)
Differentiate \(y = \ln(4x^2-1)\)
Chain rule
\(\dfrac{dy}{dx}=\dfrac{1}{4x^2-1}\cdot 8x\)
✓
\(=\dfrac{8x}{4x^2-1}\)
Connected Rates of Change
Chain rule applied to rates · Related quantities · §14.3
The chain rule for rates
When two quantities change with time, their rates are connected via the chain rule:
\(\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\)
Given the rate of change of one quantity, find the rate of change of a related quantity.
- Write the relationship between the two variables (e.g. area and radius)
- Differentiate to get \(\dfrac{dy}{dx}\)
- Multiply by the given rate \(\dfrac{dx}{dt}\)
- Substitute the given value of x
Area of circle: \(A=\pi r^2\). Radius increasing at 2 cm/s. Find rate of increase of area when r = 5.
Given
\(\dfrac{dr}{dt}=2\)
dA/dr
\(\dfrac{dA}{dr}=2\pi r\)
Chain
\(\dfrac{dA}{dt}=\dfrac{dA}{dr}\cdot\dfrac{dr}{dt}=2\pi r\cdot 2=4\pi r\)
r=5
\(\dfrac{dA}{dt}=20\pi\approx62.8\text{ cm}^2/\text{s}\)
✓
\(20\pi\) cm²/s
Small Increments & Approximations
Linear approximation · δy ≈ (dy/dx)δx · §14.4
The approximation formula
\(\delta y \approx \dfrac{dy}{dx}\cdot\delta x\)
For small changes \(\delta x\) in x, the corresponding change in y is approximately the derivative multiplied by \(\delta x\).
Percentage change: \(\dfrac{\delta y}{y}\approx\dfrac{dy/dx}{y}\cdot\delta x\). This gives the fractional (or percentage) change in y for a given fractional change in x.
\(y=x^3-2x\). Find approximate change in y when x increases from 3 to 3.02.
δx
\(\delta x = 3.02-3=0.02\)
dy/dx
\(\dfrac{dy}{dx}=3x^2-2\). At x=3: \(3(9)-2=25\)
δy
\(\delta y\approx 25\times0.02=0.5\)
✓
Approximate increase in y ≈ 0.5
Practice Questions
3 Easy · 3 Medium · 3 Hard — Applications of Differentiation
Easy 1
[3] A curve has \(y=x^3-6x^2+9x+1\). Find the intervals where the function is increasing.
SOLUTION
dy/dx
\(\frac{dy}{dx}=3x^2-12x+9=3(x-1)(x-3)\)
Increasing
\(\frac{dy}{dx}>0\): outside the roots: \(\boldsymbol{x<1}\) or \(\boldsymbol{x>3}\)
Easy 2
[3] Find the rate of change of the area of a circle with respect to its radius when \(r=5\) cm.
SOLUTION
Differentiate
\(A=\pi r^2\Rightarrow\dfrac{dA}{dr}=2\pi r\)
At r=5
\(\dfrac{dA}{dr}=10\pi\approx\boldsymbol{31.4}\) cm²/cm
Easy 3
[3] Use the small angle approximation \(\delta y\approx\frac{dy}{dx}\delta x\) to estimate the change in \(y=x^3\) when \(x\) increases from 4 to 4.02.
SOLUTION
dy/dx at x=4
\(\frac{dy}{dx}=3x^2=48\)
Estimate
\(\delta y\approx48\times0.02=\boldsymbol{0.96}\)
Medium 1
[5] A 1000 cm³ cylindrical can is to be made with minimum material. Find the dimensions (radius and height).
SOLUTION — optimisation
Constraint
\(V=\pi r^2h=1000\Rightarrow h=\dfrac{1000}{\pi r^2}\)
Surface area
\(A=2\pi r^2+2\pi rh=2\pi r^2+\dfrac{2000}{r}\)
Minimise
\(\dfrac{dA}{dr}=4\pi r-\dfrac{2000}{r^2}=0\Rightarrow r^3=\dfrac{500}{\pi}\Rightarrow r=\sqrt[3]{\dfrac{500}{\pi}}\approx5.42\) cm
\(h=\dfrac{1000}{\pi(5.42)^2}\approx10.84\) cm \(=2r\) ✓
\(h=\dfrac{1000}{\pi(5.42)^2}\approx10.84\) cm \(=2r\) ✓
Medium 2
[4] Water is poured into a conical tank at 2 cm³/s. The cone has base radius 10 cm and height 20 cm. Find the rate at which the water level rises when the depth is 8 cm.
SOLUTION — connected rates of change
Relation
Since \(r/h=10/20=1/2\Rightarrow r=h/2\)
\(V=\tfrac13\pi r^2h=\tfrac13\pi(h/2)^2h=\dfrac{\pi h^3}{12}\)
\(V=\tfrac13\pi r^2h=\tfrac13\pi(h/2)^2h=\dfrac{\pi h^3}{12}\)
Chain rule
\(\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot\dfrac{dh}{dt}\Rightarrow2=\dfrac{\pi h^2}{4}\cdot\dfrac{dh}{dt}\)
At h=8
\(\dfrac{dh}{dt}=\dfrac{2\times4}{\pi\times64}=\dfrac{8}{64\pi}=\dfrac{1}{8\pi}\approx\boldsymbol{0.0398}\) cm/s
Medium 3
[4] Find the maximum and minimum values of \(y=2\sin x+\cos 2x\) for \(0\leq x\leq\pi\).
SOLUTION
dy/dx
\(\frac{dy}{dx}=2\cos x-2\sin2x=2\cos x-4\sin x\cos x=2\cos x(1-2\sin x)=0\)
Solve
\(\cos x=0\Rightarrow x=\pi/2\) or \(\sin x=1/2\Rightarrow x=\pi/6\) or \(5\pi/6\)
Evaluate y
\(x=\pi/6: y=1+\frac12=\frac32\)
\(x=\pi/2: y=2+(-1)=1\)
\(x=5\pi/6: y=1+\frac12=\frac32\)
Max = \(\boldsymbol{\frac32}\), Min = \(\boldsymbol{1}\)
\(x=\pi/2: y=2+(-1)=1\)
\(x=5\pi/6: y=1+\frac12=\frac32\)
Max = \(\boldsymbol{\frac32}\), Min = \(\boldsymbol{1}\)
Hard 1
[6] A rectangle is inscribed in a circle of radius 5. Find the dimensions of the rectangle with maximum area.
SOLUTION
Set up
Let half-dimensions be \(x\) and \(y\). Then \(x^2+y^2=25\), so \(y=\sqrt{25-x^2}\).
Area \(A=4xy=4x\sqrt{25-x^2}\)
Area \(A=4xy=4x\sqrt{25-x^2}\)
Maximise
\(\frac{dA}{dx}=4\left(\sqrt{25-x^2}+x\cdot\frac{-x}{\sqrt{25-x^2}}\right)=4\cdot\frac{25-2x^2}{\sqrt{25-x^2}}=0\)
\(25-2x^2=0\Rightarrow x=\frac{5}{\sqrt2}\Rightarrow y=\frac{5}{\sqrt2}\)
\(25-2x^2=0\Rightarrow x=\frac{5}{\sqrt2}\Rightarrow y=\frac{5}{\sqrt2}\)
Answer
Dimensions: \(\boldsymbol{5\sqrt2\times5\sqrt2}\) (a square). Max area \(=50\) units²
Hard 2
[5] Given \(y=\dfrac{x^2}{e^x}\), find all stationary points and determine their nature. Hence sketch the curve.
SOLUTION
dy/dx
Quotient rule: \(\dfrac{2xe^x-x^2e^x}{e^{2x}}=\dfrac{x(2-x)}{e^x}=0\Rightarrow x=0\) or \(x=2\)
Points
\((0,0)\) and \((2,4/e^2)\approx(2,0.541)\)
Nature
Sign of \(\frac{dy}{dx}\): negative for \(x<0\), positive for \(02\)
\((0,0)\) is a local min; \((2,4e^{-2})\) is a local max
\((0,0)\) is a local min; \((2,4e^{-2})\) is a local max
Hard 3
[6] The volume of a cube is increasing at 3 cm³/s. Find the rate of increase of the surface area when the side length is 5 cm.
SOLUTION
V and S
\(V=x^3\Rightarrow\frac{dV}{dt}=3x^2\frac{dx}{dt}=3\Rightarrow\frac{dx}{dt}=\frac{1}{x^2}\)
\(S=6x^2\Rightarrow\frac{dS}{dt}=12x\frac{dx}{dt}=\frac{12x}{x^2}=\frac{12}{x}\)
\(S=6x^2\Rightarrow\frac{dS}{dt}=12x\frac{dx}{dt}=\frac{12x}{x^2}=\frac{12}{x}\)
At x=5
\(\dfrac{dS}{dt}=\dfrac{12}{5}=\boldsymbol{2.4}\) cm²/s
Past Year Paper Questions
Cambridge 0606 — Applications of Differentiation
0606
[10]
A closed cylindrical tin has total surface area 300\(\pi\) cm².
(a) Show that the volume is \(V=\pi r(150-r^2)\). [3]
(b) Find the value of \(r\) that maximises \(V\) and state this maximum volume. [5]
(c) Given that \(r\) is increasing at 0.1 cm/s, find the rate of increase of \(V\) when \(r=5\). [2]
(a) Show that the volume is \(V=\pi r(150-r^2)\). [3]
(b) Find the value of \(r\) that maximises \(V\) and state this maximum volume. [5]
(c) Given that \(r\) is increasing at 0.1 cm/s, find the rate of increase of \(V\) when \(r=5\). [2]
FULL SOLUTION
(a)
\(2\pi r^2+2\pi rh=300\pi\Rightarrow h=\dfrac{300\pi-2\pi r^2}{2\pi r}=\dfrac{150-r^2}{r}\)
\(V=\pi r^2h=\pi r^2\cdot\dfrac{150-r^2}{r}=\pi r(150-r^2)\) ✓
\(V=\pi r^2h=\pi r^2\cdot\dfrac{150-r^2}{r}=\pi r(150-r^2)\) ✓
(b)
\(\dfrac{dV}{dr}=\pi(150-3r^2)=0\Rightarrow r^2=50\Rightarrow r=5\sqrt2\)
\(V_{\max}=\pi\cdot5\sqrt2(150-100)=250\sqrt2\pi\approx\boldsymbol{1111}\) cm³
\(V_{\max}=\pi\cdot5\sqrt2(150-100)=250\sqrt2\pi\approx\boldsymbol{1111}\) cm³
(c)
\(\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}=\pi(150-3(25))(0.1)=\pi(75)(0.1)=\boldsymbol{7.5\pi}\approx23.6\) cm³/s
0606 Style
Rectilinear Motion with Calculus
[8]
A particle moves such that its displacement \(x\) cm at time \(t\) s is \(x=e^{2t}-4t\).
(a) Find \(v\) and \(a\) in terms of \(t\). [2]
(b) Find the minimum displacement and when it occurs. [4]
(c) Show that \(a > 0\) for all \(t \geq 0\). [2]
(a) Find \(v\) and \(a\) in terms of \(t\). [2]
(b) Find the minimum displacement and when it occurs. [4]
(c) Show that \(a > 0\) for all \(t \geq 0\). [2]
FULL WORKED SOLUTION
(a)
\(v=2e^{2t}-4\). \(a=4e^{2t}\)
(b)
\(v=0\Rightarrow 2e^{2t}=4\Rightarrow e^{2t}=2\Rightarrow t=\frac{\ln2}{2}\approx0.347\text{ s}\)
\(x=e^{\ln2}-4\cdot\frac{\ln2}{2}=2-2\ln2\approx\boldsymbol{0.614}\text{ cm}\)
\(x=e^{\ln2}-4\cdot\frac{\ln2}{2}=2-2\ln2\approx\boldsymbol{0.614}\text{ cm}\)
(c)
\(a=4e^{2t}\). Since \(e^{2t}>0\) for all \(t\), we have \(a=4e^{2t}>0\) always. ✓
0606 Style
Maclaurin / Approximation
[6]
Use the approximation \(\delta y \approx \dfrac{dy}{dx}\delta x\) with a suitable starting point to find an approximate value for \(\sqrt[3]{8.06}\), showing your working clearly. [6]
FULL WORKED SOLUTION
Let
\(y=x^{1/3}\). Start from \(x=8\) where \(y=2\) (exact). \(\delta x=0.06\)
Derivative
\(\frac{dy}{dx}=\frac{1}{3}x^{-2/3}\). At \(x=8\): \(\frac{1}{3(4)}=\frac{1}{12}\)
Approximate
\(\delta y\approx\frac{1}{12}\times0.06=0.005\)
\(\sqrt[3]{8.06}\approx2+0.005=\boldsymbol{2.005}\)
(Actual: \(\approx2.00499\)) ✓
\(\sqrt[3]{8.06}\approx2+0.005=\boldsymbol{2.005}\)
(Actual: \(\approx2.00499\)) ✓