MATHEMATICS - Institute of Education
Transcript of MATHEMATICS - Institute of Education
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 1
Name:
MATHEMATICS
Hilary Dorgan Leaving Certificate
Higher Level
2020 - 2021
C A L C U L U S
N O T E S
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 2
CONTENTS
Syllabus Page 3
Limits Page 4
Trigonometrical Limits Page 5
Differentiation from First Principles Page 5
Rules of Differentiation (Product, Quotient, Chain) Page 6
Practising the Rules of Differentiation Page 7
Differentiating Trig, Exponential & Log Functions Page 8
Differentiating Inverse Sine & Inverse Tan Functions Page 9
Second Derivatives Page 10
Applications of Calculus Page 11
Slope of a Tangent to a Curve Page 12
Finding Maxima, Minima and Points of Inflection Page 13
Curve Sketching Page 14
Displacement, Velocity, Acceleration Page 14
Rates of Change Page 16
Maxima & Minima Page 17
Sketching Derivative Curves Page 21
Implicit Differentiation Page 22
Points to Remember Page 23
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 3
Differential Calculus Paper 1 SYLLABUS
Functions and Calculus
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 4
Limits
Example 1: What is ππππβπ
(ππ + π)?
This is equivalent to asking, βwhat happens the value of 2π₯ + 3 as the value of π₯ gets closer and
closer to 5β?
Solution 1: Substitute β5β for π₯, giving an answer 2(5) + 3 = 13.
Example 2: What is ππππβπ
πβπ
ππβππ?
Solution 2: Substituting β7β for π₯, gives an indeterminate answer, 0
0. (There is no answer to this.)
But we can factorise, giving us: ππππβπ
πβπ
(πβπ)(π+π) = πππ
πβπ
π
(π+π) =
π
ππ
The laws of Limits: [Looks complicated, but all fairly obvious anyway, donβt learn off or anything , just understand]
ππππβπ
[π(π) Β± π(π)] = ππππβπ
π(π) Β± ππππβπ
π(π)
ππππβπ
[π(π). π(π)] = ππππβπ
π(π) . ππππβπ
π(π)
ππππβπ
[π(π)
π(π)] =
ππππβπ
π(π)
ππππβπ
π(π)
Example 3: What is ππππβππ
ππ β πππ
βπβπ?
Solution 3: (Substituting β25β for π₯, gives an indeterminate answer, 0
0. We need another method!)
ππππβππ
ππβ πππ
βπβπ = πππ
πβππ
(πβππ)(π+ππ) (βπ+π)
(βπβπ) (βπ+π) [Multiplying above and below by conjugate of below]
= ππππβππ
(πβππ)(π+ππ) (βπ+π)
πβππ [Denominator is equal to the difference of two squares]
= ππππβππ
(π+ππ) (βπ+π)
π =
(ππ+ππ) (βππ+π)
π = (ππ)(ππ) = πππ.
It is interesting to check limit results by substituting a value very close to the given value, e.g. 24.9 or 24.99 in
this case.
Using 24.9: ππ.ππβ πππ
βππ.πβπ =
βπ.ππ
βπ.πππππ = πππ. π. Using 24.99:
ππ.πππβ πππ
βππ.ππβπ =
βπ.ππππ
βπ.πππππππ = πππ. ππ
Example 4: What is ππππβπ
ππ
(πβπππ)β (πβπ)π ?
Solution 4: ππππβπ
ππ
(πβπππ)β (πβπ)π = ππππβπ
ππ
(πβπππ)β(πβπππ+πππβππ) = πππ
πβπ
ππ
ππβπππ = ππππβπ
π
πβπ =
βπ
π
[Note: Substituting 0.99 for x, π.πππ
(πβππ(π.ππ))β (πβπ.ππ)π = π.ππππ
βπ.πππ = βπ. ππππ β βπ. π]
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 5
Trigonometric Limits
π½ (in degrees) π½ (in radians) Sin π½ πΊππ π½
π½
10o
0.1745329252
0.1736481777 0.9949307662
1o
0.01745329252
0.01745240644
0.9999492312
0.1o
0.001745329252 0.001745328366
0.9999994923
0.01o
0.0001745329252 0.0001745329243 0.9999999949
0.001o
0.00001745329252 0.00001745329252 0.9999999999
This table shows that as the angle π (in radians) gets smaller and smaller and closer and closer to 0, then the ratio
of the Sine of the angle to the angle gets closer and closer to 1.
(This topic is also dealt with in Trigonometry notes (see page 19, Trigonometry Notes.)
Mathematically, we write this important result as: 0
sinlim 1
β= .
Differentiation from First Principles
Example 5: Find dy
dxfrom first principles if π¦ = 2π₯2 + 5π₯.
Solution 5: π(π₯) = 2π₯2 + 5π₯
π(π₯ + β) = 2(π₯ + β)2 + 5 ( π₯ + β)
= 2π₯2 + 4βπ₯ + 2β2 + 5π₯ + 5β
β π(π₯ + β) β π(π₯) = (2π₯2 + 4βπ₯ + 2β2 + 5π₯ + 5β) β (2π₯2 + 5 π₯)
= 4βπ₯ + 2β2 + 5β
βπ(π₯ + β) β π(π₯)
β = 4π₯ + 2β + 5 [Dividing both sides of equation by β]
But dy
dx =
0limhβ
π(π₯ + β) β π(π₯)
β =
0limhβ
(4π₯ + 2β + 5) = 4π₯ + 5
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 6
The Rules of Differentiation
Example 6: Prove by induction that if π(π₯) = π₯π, then ππ
ππ₯ = π π₯πβ1 for π β π.
Solution 6: (i) If π = 1: If π(π₯) = π₯1, is ππ
ππ₯ = 1 π₯1β1 = π₯0 = 1? (i.e. Is the differential of π₯ = 1?)
We know that the differential of π₯ is 1 and therefore the proposition is true for π = 1.
[But, this would have to be proved separately using first principles. Why?]
(ii) We now assume that the proposition is true for π = π.
If π = π: If π(π₯) = π₯π, then ππ
ππ₯ = ππ₯πβ1.
We must now prove that the proposition is true for π = π + 1.
If π = π + 1: To prove: If π(π₯) = π₯π+1, then ππ
ππ₯ = (π + 1) π₯π.
π(π₯) = π₯π+1 = (π₯)(π₯)π
Using product rule: LHS = ππ
ππ₯ = (π₯)(ππ₯πβ1) + (π₯)π 1 [Using proposition k, already assumed]
= ππ₯π + π₯π = π₯π (π + 1) = (π + 1)π₯π = RHS
(iii) But, proposition 1 is true and therefore, by induction, the proposition is true for all π β π.
The general rule for differentiation in example 6 above ( if π(π₯) = π₯π, then ππ
ππ₯ = π π₯πβ1 ), is in fact true for
π β π (but we only need to be able to prove it for π β π).
It is a really important rule and we use it very often to differentiate functions by rule, rather than by first
principles, which is far too tedious, unless specifically asked for.
For instance, if π(π₯) = 7 then it can also be said that π(π₯) = 7π₯0 β ππ
ππ₯ ππ πβ²(π₯) = 0 (7)(π₯)β1 = 0
Thus, the π
ππ₯ (ππππ π‘πππ‘) = 0 π΄πΏππ΄ππ.
If π(π₯) = π’(π₯) + π£(π₯), then df du dv
dx dx dx= + [The differential of a sum is the sum of the differentials of each]
If π(π₯) = π’(π₯)π£(π₯), then df dv du
u vdx dx dx
= + [The differential of a Product]
Differential of a Product = πΉπππ π‘ Γ π·ππππππππ‘πππ ππ ππππππ + ππππππ Γ π·ππππππππ‘πππ ππ πΉπππ π‘
If π(π₯) = π’(π₯)
π£(π₯), then
2
du dvv u
df dx dx
dx v
β
= [The differential of a Quotient]
Differential of a Quotient = (π΅ππ‘π‘ππ) (π·ππππππππ‘πππ ππ πππ) β (πππ) (π·ππππππππ‘πππ ππ π΅ππ‘π‘ππ)
(π΅ππ‘π‘ππ)2
The CHAIN rule of differentiation (next page): If π¦ = [π(π₯)]π β ππ¦
ππ₯ = π [π(π₯)]πβ1 [πβ²(π₯)] [Chain Rule]
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 7
PRACTISING the Rules of Differentiation
Example 7: If π¦ = π₯2(7π₯3 β 2π₯ + 3), find ππ¦
ππ₯ . [The Product Rule]
Solution 7:
π π
π π = (First) Γ (Differential of Second) + (Second) Γ (Differential of First)
= (π₯2) (21π₯2 β 2) + (7π₯3 β 2π₯ + 3)(2π₯)
= 21π₯4 β 2π₯2 + 14π₯4 β 4π₯2 + 6π₯
= 35π₯4 β 6π₯2 + 6π₯
Alternative: π¦ = π₯2(7π₯3 β 2π₯ + 3) β π¦ = 7π₯5 β 2π₯3 + 3π₯2
β ππ¦
ππ₯ = 35π₯4 β 6π₯2 + 6π₯
Example 8: If π¦ = 5π₯2 β 7π₯
2π₯ + 11 , find
ππ¦
ππ₯ . [The Quotient Rule]
Solution 8:
π π
π π =
(π΅ππ‘π‘ππ) (π·ππππππππ‘πππ ππ πππ) β (πππ) (π·ππππππππ‘πππ ππ π΅ππ‘π‘ππ)
(π΅ππ‘π‘ππ)2
= (2π₯ + 11) (10π₯ β 7) β (5π₯2 β 7π₯) (2)
(2π₯ + 11)2 = 20π₯2 β 14π₯ + 110π₯ β 77 β 10π₯2 + 14π₯
(2π₯ + 11)2
= 10π₯2 + 110π₯ β 77
(2π₯ + 11)2
The CHAIN rule of differentiation: If π¦ = [π(π₯)]π β ππ¦
ππ₯ = π [π(π₯)]πβ1 [πβ²(π₯)]
Example 9: Differentiate (i) β7π₯ β 11 , (ii) βπ₯2β1
2π₯+1
3
Solution 9: (i) π(π₯) = β7π₯ β 11 = (7π₯ β 11)1
2 [Always re-write question in calculus friendly form]
β πβ² (π₯) = 1
2 (7π₯ β 11)
β1
2 (7) [β7β is the differential of what is in the brackets]
= 7
2 β7π₯β11
(ii) π(π₯) = βπ₯2β 1
6π₯ + 1
3 = (
π₯2β 1
6π₯ + 1 )
1
3 [Always re-write question in calculus friendly form]
β πβ² (π₯) = 1
3 (
π₯2β1
6π₯+1)
β2
3 Γ (differential of what is in the brackets)
= 1
3 (
π₯2β 1
6π₯+ 1)
β2
3 [
(6π₯+1) (2π₯)β(π₯2β1)(6)
(6π₯+1)2 ]
= 1
3 (
π₯2β 1
6π₯+ 1)
β2
3 [
12π₯2 + 2π₯ β 6π₯2+ 6
(6π₯+1)2 ]
= 1
3 (
π₯2β 1
6π₯ + 1)
β2
3 [
6π₯2+2π₯+6
(6π₯ + 1)2 ] = 2
3 (
π₯2β 1
6π₯ + 1)
β2
3 [
3π₯2 + π₯ + 3
(6π₯ + 1)2 ]
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 8
Differentiating trigonometric functions
Example 10: If π¦ = π ππ (π₯3 β 5π₯2 + 3), find ππ¦
ππ₯.
Solution 10: ππ¦
ππ₯= πππ (π₯3 β 5π₯2 + 3) (3π₯2 β 10π₯) = (3π₯2 β 10π₯) πππ (π₯3 β 5π₯2 + 3).
Example 11: If π¦ = 1βπππ 2π₯
7+ π ππ 3π₯, find
ππ¦
ππ₯ at π₯ = 0.
Solution 11: ππ¦
ππ₯=
(7+ π ππ 3π₯)(2 π ππ 2π₯) β (1 β πππ 2π₯) (3 πππ 3π₯)
(7 + π ππ 3π₯)2
ππ¦
ππ₯ at π₯ = 0:
(7 + 0)(0) β (0)(3)
49 = 0
Example 12: If π(π₯) = π‘ππ5π₯
3π₯β1 find πβ²(π₯) at π₯ = 1.
Solution 12: πβ²(π₯) = π ππ2 (5π₯
3π₯β1 ) Γ
(3π₯ β 1) (5) β (5π₯) (3)
(3π₯ β 1)2
β πβ²(1) = π ππ2 (5
3 β 1 ) Γ
(3β1) (5) β (5) (3)
(3 β 1)2 = π ππ2 (5
2 ) Γ
10 β 15
4 =
β5
4 π ππ2 (
5
2) β β0.80
Differentiating exponential functions
Example 13: If π(π₯) = 17πβ2π₯2 find πβ²(π₯).
Solution 13: πβ²(π₯) = (17 πβ2π₯2 ) Γ (β4π₯) = β68π₯πβ2π₯2 .
Example 14: If π¦ = π₯5 πβ3π₯ find ππ¦
ππ₯ at π₯ = 2.
Solution 14: ππ¦
ππ₯= (π₯5) (πβ3π₯) (β3) + (πβ3π₯) (5π₯4) = πβ3π₯(β3π₯5 + 5π₯4)
ππ¦
ππ₯ at π₯ = 2: πβ6(β3(2)5 + 5(2)4) = πβ6(β96 + 80) = β16
π6β
Example 15: Differentiate π2π₯
3π₯2β with respect to π₯.
Solution 15: πβ²(π₯) = (3π₯2) (2 π2π₯)β (π2π₯)(6π₯)
(3π₯2)2 = (6π₯2β 6π₯)(π2π₯)
9π₯4
Differentiating logarithmic functions
Example 16: If π¦ = ππππ(2 π ππ 3π₯), find ππ¦
ππ₯ at π₯ =
π
12.
Solution 16: ππ¦
ππ₯=
πβ²(π₯)
π(π₯) =
2 (3) πππ 3π₯
2 π ππ 3π₯ =
3
π‘ππ 3π₯
ππ¦
ππ₯ at π₯ =
π
12:
3
π‘ππ π
4
= 3
1 = 3
Example 17: If π(π₯) = [ππ (13π₯ β 5)]3 , find (i) πβ²(π₯) and (ii) πβ²(2) correct to two places of decimal.
Solution 17: (i) πβ²(π₯) = 3 [ππ (13π₯ β 5)]2 (13
(13π₯β5)) =
39 [ππ (13π₯β5)]2
13π₯β5
(ii) πβ²(2) = 39 [ππ (26β5)]2
26β5 =
39 [ππ (21)]2
21 =
13 (3.044522)2
21 =
13 (9.26912)
21 β 5.74
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 9
Example 18: Differentiate ππππ β2π₯2
π₯2+1
Solution 18: π¦ = ππππ β2π₯2
π₯2+1 β π¦ =
1
2 ππππ
2π₯2
π₯2+1 =
1
2 [ππππ(2π₯2) β ππππ(π₯2 + 1)]
[Using rules of logs]
ππ¦
ππ₯ =
1
2 [
4π₯
2π₯2 β 2π₯
π₯2+1 ] =
1
2 [
2
π₯ β
2π₯
π₯2+1 ] =
1
π₯ β
π₯
π₯2+1
= [ (1)(π₯2+1) β (π₯)(π₯)
(π₯) (π₯2+1) ]
= (π₯2+ 1) β (π₯2)
(π₯) (π₯2+1)
= 1
(π₯) (π₯2+1)
Differentiating inverse trigonometric functions like πππβπ π and πππβπ π
Page 25 of the Mathematical Tables tell us that if π(π₯) = π ππβ1 π₯
π then πβ²(π₯) =
1
βπ2β π₯2 , where π is a constant.
It also tells us that if π(π₯) = π‘ππβ1 π₯
π then πβ²(π₯) =
1
π2 + π₯2 , where π is a constant.
π
ππ₯ (π ππβ1
π₯
π) =
1
βπ2β π₯2
π
ππ₯ (π‘ππβ1
π₯
π) =
π
π2 + π₯2
If π¦ = π ππβ1 π(π₯), ππ¦
ππ₯=
1
β1β[π(π₯)]2 Γ πβ²(π₯) AND If π¦ = π‘ππβ1 π(π₯),
ππ¦
ππ₯=
1
1 + [π(π₯)]2 Γ πβ²(π₯)
Example 19: If (π₯) = π‘ππβ1 π₯
3 , find πβ²(π₯).
Solution 19: Method 1: πβ²(π₯) = 3
32 + π₯2 = 3
9 + π₯2 [Because π₯ has been replaced by β3β]
Method 2: πβ²(π₯) = 1
1 +( π₯
3)2
Γ 1
3 =
9
9 + π₯2 Γ 1
3=
3
9 + π₯2
Example 20: If (π₯) = (π ππβ1 3π₯)5 , find πβ²(π₯).
Solution 20: πβ²(π₯) = 5(π ππβ1 3π₯)4 Γ 1
β1β (3π₯)2 Γ 3 =
15(π ππβ1 3π₯)4
β1β (3π₯)2
Example 21: If (π₯) = π‘ππβ1 ( 3π₯
2π₯β7 ) , show that πβ²(0) =
β 3
7 .
Solution 21: πβ²(π₯) = 1
1+(3π₯
2π₯β7)2
Γ (2π₯β7)(3)β(3π₯)(2)
(2π₯β7)2 = (2π₯β7)2
(2π₯β7)2+(3π₯)2 Γ 6π₯β21β6π₯
(2π₯β7)2
= β21
(2π₯β7)2+(3π₯)2 = β21
4π₯2β28π₯+49+9π₯2
= β21
13π₯2β28π₯+49
β πβ²(0) = β21
49 =
β 3
7
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 10
Second Derivatives
If π¦ = π(π₯), then π2π¦
ππ₯2 = π
ππ₯ (
ππ¦
ππ₯ ) = πβ²β² (π₯).
Example 22: If π¦ = π4π₯, show that π2π¦
ππ₯2 β 8 ππ¦
ππ₯ + 16π¦ = 0.
Solution 22: ππ¦
ππ₯ =
π
ππ₯(π4π₯) = 4π4π₯
β π2π¦
ππ₯2 = π
ππ₯ (
ππ¦
ππ₯ ) =
π
ππ₯(4π4π₯ ) = 16π4π₯.
β΄ πΏπ»π = π2π¦
ππ₯2 β 8 ππ¦
ππ₯ + 16π¦
= 16π4π₯ β (8)(4π4π₯) + 16 (π4π₯)
= π4π₯(16 β 32 + 16)
= (π4π₯) 0 = 0 = π π»π
Example 23: Given that π¦ = π₯π3π₯, find the value of (ππ¦
ππ₯)
2β π3π₯
π2π¦
ππ₯2 when π₯ = 0.
Solution 23: π¦ = π₯π3π₯ β ππ¦
ππ₯ = (π₯)(3π3π₯) + (π3π₯)(1)
= π3π₯(3π₯ + 1)
β π2π¦
ππ₯2 = π
ππ₯ [π3π₯(3π₯ + 1)]
= (π3π₯)(3) + (3π₯ + 1) (3π3π₯)
= (3π3π₯) (1 + 3π₯ + 1)
= (3π3π₯) (3π₯ + 2)
β΄ (ππ¦
ππ₯)
2
β π3π₯ π2π¦
ππ₯2 = [π3π₯(3π₯ + 1)]2 β (π3π₯)(3π3π₯)(3π₯ + 2)
β΄ (ππ¦
ππ₯)
2β π3π₯
π2π¦
ππ₯2 at π₯ = 0:
= [1(1)]2 β (1)(3)(2)
= 1 β 6 = β5
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 11
Finding the slope and equation of a tangent to a curve
A vital application of differentiation is the following:
ππ¦
ππ₯ is the slope of a tangent to a curve at any point on the curve.
To find the equation of a tangent to a curve at a given point (π₯1, π¦1) on the curve, take the following steps:
1) Find ππ¦
ππ₯
2) Evaluate ππ¦
ππ₯ at π₯ = π₯1 [This gives π, the slope of the tangent to the curve at (π₯1, π¦1) ]
3) Use equation π¦ β π¦1 = π (π₯ β π₯1) to find the equation of the tangent to the curve at (π₯1, π¦1)
Example 24: Find the equation of the tangent to the curve π¦ = 10 β πππ π₯ at the point ( π
2 , 0 )
Solution 24: π¦ = 10 β πππ π₯
β ππ¦
ππ₯ = 0 + π ππ π₯
ππ¦
ππ₯ @π₯ =
π
2 = π ππ
π
2 = 1 [This gives π, the slope of the tangent to the curve at (
π
2 , 0 )]
β Eqn of tangent: π¦ β π¦1 = π (π₯ β π₯1)
β Eqn of tangent: π¦ β 0 = 1 (π₯ β π
2 ) β π¦ = π₯ β
π
2
Example 25: The equation of a curve is π¦ = ππ₯3 + ππ₯2 + ππ₯ + π, where π, π, π, π β π.
The curve passes through the points (0, 15) and (2, β15) and at these points the slopes of the
tangents are both β13.
(i) Find the values of π, π, π, and π
(ii) Show that the graph cuts the π₯ β ππ₯ππ at the point π₯ = 5.
(iii) Find the coordinates of the other two points where the curve cuts the π₯ β ππ₯ππ .
(iv) Hence, sketch the curve.
Solution 25: (i) (0, 15) β π¦ = ππ₯3 + ππ₯2 + ππ₯ + π β 15 = 0 + 0 + 0 + π β π = ππ
(2, β15) β π¦ = ππ₯3 + ππ₯2 + ππ₯ + π β β15 = 8π + 4π + 2π + 15
β 8π + 4π + 2π = β30
β 4π + 2π + π = β15 β¦ eqn 1
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 12
ππ¦
ππ₯ = 3ππ₯2 + 2ππ₯ + π
ππ¦
ππ₯ @π₯ = 0: 0 + 0 + π = β13 β π = βππ
ππ¦
ππ₯ @π₯ = 2: 3π(2)2 + 2π(2) + (β13) = 12π + 4π + (β13) = β13
β 12π + 4π = 0 β 3π + π = 0 β¦ eqn 2
From eqn 1: 4π + 2π β 13 = β15 β 2π + π = β1
Using 3π + π = 0 and 2π + π = β1
β βπ = βπ β π = π and π = βπ
(ii) If π = π: π¦ = ππ₯3 + ππ₯2 + ππ₯ + π β π¦ = 1π₯3 β 3π₯2 β 13π₯ + 15
β If π = π: β π¦ = 1(5)3 β 3(5)2 β 13(5) + 15 = 125 β 75 β 65 + 15 = 140 β 140 = 0
β the graph cuts the π₯ β ππ₯ππ at (5, 0).
(iii) π(π) = π β π β π is a factor
(π₯3 β 3π₯2 β 13π₯ + 15) Γ· (π₯ β 5) = π₯2 + 2π₯ β 3 = (π₯ β 1) (π₯ + 3)
β other two points where the curve cuts the π₯ β ππ₯ππ βΆ (1, 0) and (β3, 0).
(iv) Sketch
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 13
Finding maxima, minima and points of inflection of a curve
Local Maximum At local maximum: ππ¦
ππ₯= 0 and
π2π¦
ππ₯2 < 0.
At local minimum: ππ¦
ππ₯= 0 and
π2π¦
ππ₯2 > 0.
At point of inflection: π2π¦
ππ₯2 = 0.
Point of Inflection Curves are increasing when ππ¦
ππ₯> 0.
Curves are decreasing when ππ¦
ππ₯< 0.
At turning points, the slope of the tangent to
the curve is parallel to the x-axis, i.e. ππ¦
ππ₯= 0.
Local Minimum
Example 26: Find the local maximum, the local minimum and a point of inflection of the curve
π¦ = β π₯3 + 6π₯2 β 9π₯ + 3. Use this information to sketch the curve.
Solution 26: If π¦ = β π₯3 + 6π₯2 β 9π₯ + 3
β ππ¦
ππ₯ = β3π₯2 + 12π₯ β 9 Local Max (3, 3)
To find turning points, we must solve ππ¦
ππ₯= 0
β΄ β3π₯2 + 12π₯ β 9 = 0 β π₯2 β 4π₯ + 3 = 0
β΄ (π₯ β 3)(π₯ β 1) = 0
β π₯ = 3 ππ π₯ = 1 β Turning points @ (3, 3) & (1, β1) Point of Inflection (2, 1)
π2π¦
ππ₯2 = β6π₯ + 12 [π¦ coordinate is calculated from original equation]
π2π¦
ππ₯2 @ π₯ = 3: β 6(3) + 12 = β6 < 0 β Max (3, 3)
π2π¦
ππ₯2 @ π₯ = 1: β 6(1) + 12 = + 6 > 0 β Min (1, β1)
If π2π¦
ππ₯2 = 0: β β6π₯ + 12 = 0 β π₯ = 2 , π¦ = 1 Local Min (1, β 1)
β΄ Point of inflection @ (2, 1)
Example 27: Find the coordinates of the one turning point of the curve π¦ = β π2π₯ + 2 ππ₯ and determine
whether this turning point is a maximum or a minimum.
Solution 27: If π¦ = β π2π₯ + 2ππ₯ , then ππ¦
ππ₯= β 2π2π₯ + 2ππ₯.
Turning point when ππ¦
ππ₯= β 2π2π₯ + 2ππ₯ = 0 β 2π2π₯ β 2ππ₯ = 0.
β π2π₯ β ππ₯ = 0 β ππ₯ (ππ₯ β 1) = 0
β ππ₯ = 0 or ππ₯ = 1
ππ₯ = 0 has no solution and if ππ₯ = 1, then π₯ = 0.
β΄ Turning point at (0, 1) [If π₯ = 0, π¦ = β π2(0) + 2π(0) = β1 + 2 = 1]
π2π¦
ππ₯2 = β4 π2π₯ + 2ππ₯
β΄ π2π¦
ππ₯2 @ π₯ = 0: β 4 π2(0) + 2π(0) = β4 + 2 = β2 < 0 β turning point (0, 1) is a Maximum.
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 14
Sketching Curves with the help of calculus
Example 28: π¦ = π₯ β 1
π₯ β 2 is a curve.
(i) Find the equation of the vertical asymptote of this curve.
(ii) Find the equation of the horizontal asymptote of this curve.
(iii) Prove that the curve has no turning points
(iv) Sketch the curve.
(v) If the slopes of the tangents to this curve at (π₯1, π¦1) and (π₯2, π¦2) are parallel and π₯1 β
π₯2, prove that π₯1 + π₯2 = 4 .
Solution 28:
(i) Vertical asymptote is found by letting the denominator = 0.
β΄ π₯ β 2 = 0 β π = π [This is the equation of vertical asymptote].
(ii) Horizontal asymptote is found by letting π¦ = ππππ₯ββ
π(π₯).
π¦ = ππππ₯ββ
π(π₯) β π¦ = ππππ₯ββ
π₯β1
π₯β2 .
To find the limit as π₯ β β, divide above and below by highest power of π₯.
β΄ π¦ = ππππ₯ββ
1 β 1π₯β
1 β 2 π₯β =
1 β 0
1 β 0 = 1
β΄ Equation of horizontal asymptote: π = π.
(iii) ππ¦
ππ₯ =
(π₯β2)(1)β(π₯β1)(1)
(π₯β2)2 = β 1
(π₯β2)2
There is no value of π₯ which will make ππ¦
ππ₯= 0 β No turning points. asymptotes
(iv) Find coordinates of points to the left and right of π₯ = 2.
(0, 1
2 ), (1, 0) (3, 2) and (4,
3
2 ) are on the curve. [Graph paper should be used when sketching the curve (see graph)]
(v) The slope (π1)of the tangent to the curve at (π₯1, π¦1) = πβ²(π₯1). But, πβ²(π₯1) = β 1
(π₯1β2)2 .
The slope (π2) of the tangent to the curve at (π₯2, π¦2) = πβ²(π₯2). But, πβ²(π₯2) = β 1
(π₯2β2)2 .
But, (π1) = (π2) β (π₯1 β 2)2 = (π₯2 β 2)2 [But, π₯1 β π₯2 β (π₯1 β 2) β (π₯2 β 2)]
β΄ π₯1 β 2 = β(π₯2 β 2) β π₯1 + π₯2 = 4.
Displacement, Velocity and Acceleration
We often use the letter π to represent displacement (distance travelled in a certain direction).
Velocity (π£)is the rate of change of displacement with respect to time.
In calculus, we can write the previous sentence as π£ = ππ
ππ‘ .
Similarly, acceleration (π) is the rate of change of velocity with respect to time.
In calculus, we can write the previous sentence as π = ππ£
ππ‘ .
Thus, π =ππ£
ππ‘ =
π2π
π π‘2 or, in English, acceleration is the second differential of displacement with respect to time.
Note: In calculus, if the βrate of changeβ of something is mentioned and there is no βwith respect to β¦β, then it can
always be assumed that it is with respect to βtimeβ. That is, βtimeβ is the default if nothing else is mentioned.
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 15
Example 29: An object moves in a straight line such that its distance from a fixed point, π΄ , is given by
s = π‘3 β 12π‘2 + 21π‘ + 5 , where π is in metres and π is in seconds.
(i) Find the displacement from the point π΄ at the start of the motion.
(ii) Find the greatest displacement of the object from π΄ (i.e. find local maximum)
(iii) When does the object come to instantaneous rest?
(iv) Find the speed of the object when π‘ = 10 seconds.
(v) Find the minimum velocity of the object.
(vi) Find the acceleration of the particle after five seconds.
Solution 29:
(i) Finding π , @ π‘ = 0:
s = π‘3 β 12π‘2 + 21π‘ + 5
β΄ π @ π‘ = 0: = (0)3 β 12(0)2 + 21(0) + 5 = 5 m
(ii) Displacement is greatest when ππ
ππ‘= 0
ππ
ππ‘= 3π‘2 β 24π‘ + 21 = 0
β π‘2 β 8π‘ + 7 = 0
β (π‘ β 7) (π‘ β 1) = 0
β π‘ = 7 or π‘ = 1 s
Note: π2π
ππ‘2 = 6π‘ β 24
At π‘ = 7: π2π
ππ‘2 = 6(7) β 24 = 18 m/s > 0 β Minimum displacement at π‘ = 7
At π‘ = 1: π2π
ππ‘2 = 6(1) β 24 = β18 m/s < 0 β Maximum displacement at π‘ = 1
When π‘ = 1, π = (1)3 β 12(1)2 + 21(1) + 5 = 15 m
(iii) Instantaneous rest β π£ = 0.
As in part (ii) above, π£ = ππ
ππ‘= 0 when π‘ = 1 and when π‘ = 7 s.
(iv) Finding π£, @ π‘ = 10
π£ = ππ
ππ‘ = 3π‘2 β 24π‘ + 21
β π£, @ π‘ = 10: 3(10)2 β 24(10) + 21 = 81 m/s
(v) Velocity is minimum (or lowest) when ππ£
ππ‘ =
π2π
ππ‘2 = 6π‘ β 24 = 0
β Velocity is minimum when π‘ = 4 s
[obviously a minimum rather than a maximum here as the second differential = +6 > 0]
π£ = 3π‘2 β 24π‘ + 21
β π£πππ = 3(4)2 β 24(4) + 21 = β27 m/s
(vi) Acceleration (π) = ππ£
ππ‘ = 6π‘ β 24
β π , @ π‘ = 5 = 6(5) β 24
β π = 6 m/s2
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 16
Rates of Change
There are many examples in life where one quantity changes with another. The rate of how one quantity (like sale
of ice creams) varies with another (like temperature) is a very important application of Calculus.
Many problems on rates of change involve displacement, velocity and acceleration that we have seen above.
Velocity (or speed) is the rate of change of displacement, i.e. π£ = ππ
ππ‘ . It is measured in m/s.
Acceleration is the rate of change of velocity, i.e. π = ππ£
ππ‘ =
π2π
ππ‘2 . It is measured in m/s2.
Example 30: A ball is fired straight up in the air. The height, β, of the ball above the ground is given
by β = 30π‘ β 5π‘2 where π‘ is the time in seconds after the ball was fired.
(i) After how many seconds does the ball hit the ground?
(ii) Find the speed of the ball after 2 seconds.
(iii) Find the maximum height reached by the ball.
(iv) Find the constant deceleration of the ball.
(v) When is the speed of the ball equal to 15 m/s?
Solution 30: (i) When the ball hits the ground, β = 0.
β = 30π‘ β 5π‘2 β 0 = 30π‘ β 5π‘2 β 5π‘2 β 30 π‘ = 0
β 5π‘ (π‘ β 6) = 0 β π‘ = 0 ππ π‘ = 6 seconds Answer: π = π seconds
(ii) Speed of ball at any time = π£ = πβ
ππ‘ = 30 β 10 π‘
β΄ Speed of the ball after 2 seconds, π£ or πβ
ππ‘ at π‘ = 2: 30 β 10(2) = 30 β 20 = ππ m/s.
(iii) Height is maximum when π£ or πβ
ππ‘ = 0 , i.e. 30 β 10 π‘ = 0 β π‘ = 3 seconds
Maximum height = β at π‘ = 3: β = 30π‘ β 5π‘2
= 30(3) β 5(3)2 = 90 β 45 = ππ m
(iv) Acceleration = ππ£
ππ‘ or
π2π
ππ‘2 = π
ππ‘(30 β 10 π‘) = β10 m/s2
β΄ Deceleration = ππ m/s2.
(v) Speed of ball at any time = π£ = πβ
ππ‘ = 30 β 10 π‘
Speed of ball = 15 m/s when = π£ = πβ
ππ‘ = 30 β 10 π‘ = 15
β 15 = 10 π‘ β π‘ = π. π s
Example 31: A stone was dropped in a pond. The radius of a circle (or wave) produced increased at a rate of 75 cm/s.
Find the rate at which the area inside the circle increased when the radius of the circle was 5 m.
Solution 31: Given: ππ
ππ‘ = 0.75 [We cannot have a mixture of units, so bring lengths to metres]
Required to Find: ππ΄
ππ‘ @ π = 5
Express one variable in terms of the other: π΄ = π π2 [Formula for area of a disc/circle]
β΄π π¨
π π =
ππ΄
ππ Γ
ππ
ππ‘ = 2ππ Γ 0.75 [Differentiating using chain rule and
π
ππ(π π2) = 2ππ ]
β΄π π¨
π π @ π = 5: 1.5 π(5) =
15π
2 m2/s.
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 17
Example 32: A large spherical balloon is being blown up and its radius is increasing at the rate of 5 cm/s.
When the radius of the balloon is 30 cm, find the rate of increase of (i) its volume and (ii) its surface area.
Solution 32: (i) Given: ππ
ππ‘ = 5 [All lengths are in cm, so there is no need to convert to metres]
Required to Find: ππ
ππ‘ @ π = 30
Express one variable in terms of the other: π =4
3 π π3 [Formula for volume of a sphere]
β΄π π½
π π =
ππ
ππ Γ
ππ
ππ‘ = 4ππ2 Γ 5 = 20ππ2 . [Differentiating using chain rule and
π
ππ(
4
3 π π3) = 4π π2]
β΄π π½
π π @ π = 30: 20 π(30)2 = 18 000 π cm3/s.
(ii) Given: ππ
ππ‘ = 5 [All lengths are in cm, so there is no need to convert to metres]
Required to Find: ππ΄
ππ‘ @ π = 30
Express one variable in terms of the other: π΄ = 4 π π2 [Formula for surface area of a sphere]
β΄π π¨
π π =
ππ΄
ππ Γ
ππ
ππ‘ = 8ππ Γ 5 = 40ππ . [Differentiating using chain rule and
π
ππ(4 π π2) = 8ππ]
β΄π π¨
π π @ π = 30: 40 π (30) = 1200 π cm2/s.
Maximum and Minimum Problems
The procedure for tackling maximum and minimum problems can be summarized as follows:
(1) Draw a diagram representing the problem (if appropriate) and label all the relevant details.
(2) Find a formula or equation for the quantity that is to be maximized or minimized.
(3) Express the quantity to be maximized (or minimized) in terms of one variable, using the conditions given
in the problem, if necessary, to eliminate any other variable(s).
(4) Differentiate once to find where ππ¦
ππ₯= 0 (i.e. maximum or minimum point).
(5) Differentiate a second time to distinguish between the maximum and the minimum.
Example 33: A farmer wishes to enclose a rectangular section in the middle of a field.
She uses all of πππ m of an electric fence to construct the enclosure.
(a) If one side of the rectangular enclosure is π metres long, what is the length of the adjacent side?
(b) Find, in terms of π, the area of the enclosure.
(c) Find the value of π that makes the area of the enclosure a maximum.
(d) If the farmer constructs a circular enclosure instead of a rectangular one (as above), find the radius
of the enclosure correct to 2 decimal places, assuming she uses all πππ m of the electric fence to
form the enclosure.
(e) Find the area of the circular enclosure to the nearest square metre.
(f) If the farmer wants the area of the enclosure to be a maximum (using exactly πππ m of electric
fencing), should she construct a circular or a rectangular enclosure?
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 18
Solution 33: (a) π¦
π¦ = 1
2 (400 β 2π₯)
π₯ = 200 β π₯
[using info given to express one variable in terms of the other]
(b) To be maximized: Area(A) = length Γ breadth = π₯(200 β π₯) = 200π₯ β π₯2
(c) π π¨
π π = πππ β ππ , which is a maximum when
π π¨
π π = πππ β ππ = π β π = πππ.
π2π΄
ππ₯2 = β2 < 0 β Maximum @ π₯ = 100
(d) circumference length= 2ππ β 400 = 2ππ
β π = 400
2π =
200
π = 63.66 m π
(e) Area = ππ2 = π (63.66)2 = 12 732 m2
(f) Maximum rectangular area = 100 Γ 100 = 10 000 m2
β biggest area formed by circle
Example 34: If two natural numbers add to give ππ, find the maximum product of these two numbers. Solution 34: Let the two numbers be π₯ and π¦. But, π¦ = 30 β π₯ [using info given to express one variable in terms of the other] To be maximized: Product (π) = π₯ Γ (30 β π₯) = 30π₯ β π₯2
ππ
ππ₯ = 30 β 2π₯ , which is a maximum when
ππ
ππ₯ = 30 β 2π₯ = 0 β π₯ = 15.
π2π΄
ππ₯2 = β2 < 0 β Maximum @ π₯ = 15
Maximum product π₯π¦ = 15(30 β 15) = 225
Example 35: A cylindrical can is used to hold πππ cm3 of orange drink. π
(i) Find the height (β) of the can in terms of the radius (π).
(ii) Find the dimensions of the can that will use the least amount
of metal in its manufacture.
β
Solution 35: (i) Volume of cylinder= 330 = ππ2β β β = 330
ππ2
(ii) To be minimized: Surface Area (π΄) = 2ππβ + 2(ππ2)
= 2ππ330
ππ2 + 2ππ2
= 660
π + 2ππ2 = 660πβ1 + 2ππ2
β΄ ππ΄
ππ = β660 πβ2 + 4ππ, which is a maximum when
ππ΄
ππ = β660 πβ2 + 4ππ = 0
β 660
π2 = 4ππ
β 660 = 4ππ3
β π = β660
4π
3 = β
165
π
3 = 3.745 cm
Note: π2π΄
ππ2 = 1320
π3 + 4π which is > 0 when π = 3.745 β Minimum @ π = 3.745.
When π = 3.745: β = 330
ππ2 = 330
π(3.745)2 = 330
44.06 = 7.49 cm
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 19
Example 36: An equilateral triangle of length π cm has a rectangle inscribed in it.
(a) Find β, the height of the triangle. π cm β
(b) Show that π = βπ π.
(c) Find, in terms of π the area of the rectangle. π¦
(d) Hence, find the maximum possible area of this rectangle.
(e) What percentage of the triangle is occupied by the maximum π₯ π₯
sized rectangle? 8
Solution 36: (a) By Pythagorasβ theorem: β2 = 82 β 42 = 64 β 16 = 48
β β = β48 ππ 4β3 cm
(b) π‘ππ 60π = π¦
π₯ β β3 =
π¦
π₯ β π¦ = β3 π₯
(c) Area of Rectangle (π΄) = (8 β 2π₯) (π¦)
= (8 β 2π₯) (β3 π₯)
= 8 β3 π₯ β 2β3 π₯2 cm2
(d) ππ΄
ππ₯ = 8 β3 β 4β3 π₯,
β΄ Area is a maximum when ππ΄
ππ₯ = 8 β3 β 4β3 π₯ = π
β π₯ = 8 β3
4 β3 = 2
But, π2π΄
ππ₯2 = β 4β3 < 0 β Maximum @ π₯ = 2
β΄ π΄πππ₯ = 8 β3 (2) β 2β3 (2)2
= 16 β3 β 8 β3
= 8 β3 cm2
(e) Area of Triangle = 1
2 (8)(4β3)
= 16 β3 cm2
β΄ % Area occupied by maximum rectangle = 8 β3
16 β3 Γ
100
1
= 50%
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 20
Example 37: π¨π©πͺπ« is a rectangular ploughed field πππ m long and πππ m wide with a path around its sides.
Tom can run at a speed of πππβ m/s along the path
and at a speed of π πβ m/s across the ploughed field. π« 310 πͺ
Tom wants to reach point πͺ from the point π¨ as quickly as possible.
He runs from π¨ to a point π¬ along the path and from π¬ to πͺ across 120
the field.
(a) If |π¬π©| = π, express in terms of π : π¨ π¬ π π©
(i) |π¨π¬|
(ii) |π¬πͺ|
(iii) the total time taken to travel from π¨ to πͺ
(b) Find the value of π for which the time is a minimum.
(c) Calculate the minimum time to travel from π¨ to πͺ (in seconds).
Solution 37: (a) (i) |π¨π¬| = 310 β π₯
(ii) |π¬πͺ|2 = π₯2 + 1202
β |π¬πͺ| = βπ₯2 + 1202
(iii) time taken = πππ π‘ππππ
π ππππ
= (3) (310 β π₯)
13 +
(3)βπ₯2 + 1202
5
(b) To be minimized: time = π‘ = (3) (310 β π₯)
13 +
(3)βπ₯2+ 1202
5
β π π
π π =
βπ
ππ +
π
π π
π (ππ + 1202)
βπ
π (ππ) = π [when time is a minimum]
β π
ππ =
π
π
π
βπ₯2+ 1202
β πππ = π βπ₯2 + 1202
β πππππ = ππ (ππ + 1202)
β πππππ = ππ (1202)
β πππ = Β± π (120)
β π = Β± π (10) = Β± ππ
β π = ππ
β΄ time is a minimum when π = ππ m
(c) Minimum time = (3) (310 β π₯)
13 +
(3)βπ₯2+ 1202
5
= (3) (310 β 50)
13 +
(3)β502+ 1202
5
= 60 + (3)130
5
= 60 + 78
= 138 s
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 21
Sketching Derivative Curves
If we are given the graph of a curve π¦ = π(π₯), we can sketch the graph of its derivative, ππ¦
ππ₯ ππ πβ²(π₯),
by examining the sign of the derivative and how the slope of the tangent is changing.
The diagram below shows a cubic function π¦ = π(π₯), with turning points at (0, 8) and (1.1, 6).
Also shown are the functions πβ²(π₯) and πβ²β²(π₯).
Note: 1) At a maximum or minimum of one curve, the derivative curve will cut the π₯-axis
i.e. ππ¦
ππ₯= 0 at a maximum and at a minimum
2) At a point of inflection of one curve, the derivative curve will have a turning point
and the second derivative curve will cut the π₯ βaxis.
i.e. ππ¦
ππ₯ = 0 and
π2π¦
ππ₯2 = 0 at a point of inflection
3) After that, check whether the slope of the original is positive or negative
and whether it is increasing or decreasing.
Minimum at (1.1, 6) β πβ²(1.1) = 0 β (1.1, 0) is on the derivative curve πβ²(π₯).
Maximum at (0, 8) β πβ²(0) = 0 β (0, 0) is on the derivative curve πβ²(π₯).
A point of inflection at π₯ = 0.55 on π¦ = π(π₯) β turning point at π₯ = 0.55 on πβ²(π₯)
AND (0.55, 0) is on the second derivative curve πβ²β²(π₯)
Maximum at π₯ = 0 on π¦ = π(π₯) β (0, 0) is on the derivative curve, π¦ = πβ²(π₯).
Slope of π¦ = πβ²(π₯) is negative but decreasing negatively to the left of π₯ = 0.55, but is positive and increasing
to the right of π₯ = 0.55 β π¦ = πβ²β²(π₯) has a positive slope, cutting π₯-axis at (0.55, 0).
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 22
Implicit Differentiation (2015+)
[Syllabus Addition (2015): Use differentiation to find the slope of a tangent to a circle]
It is easy to find π (π₯2)
ππ₯ . The answer is 2π₯. But, what if we are asked to find
π (π¦2)
ππ₯ .
The technique for doing this is known as implicit differentiation. We have to use the Chain Rule to help us out.
π (π¦2)
ππ₯=
π π¦2
ππ¦ ππ¦
ππ₯ ( i.e. multiply above and below by ππ¦) (not forgetting the sideways kick). The answer is: 2π¦
ππ¦
ππ₯ .
Examples: (i) π (7π¦3)
ππ₯ =
π (7π¦3)
ππ¦ ππ¦
ππ₯ = 21π¦2
ππ¦
ππ₯
(ii) π (βπ¦)
ππ₯ =
π (βπ¦)
ππ¦ ππ¦
ππ₯ =
1
2π¦β1/2
ππ¦
ππ₯ =
1
2βπ¦ ππ¦
ππ₯
Example 38: A circle has an equation π₯2 + π¦2 β 4π₯ + 2π¦ β 20 = 0 .
Using differentiation, find the equation of the tangent to this circle at the point (5, β5).
Solution38:
Step 1: Differentiate both sides of the equation with respect to π₯.
2π₯ + 2π¦ ππ¦
ππ₯ β 4 + 2
ππ¦
ππ₯ β 0 = 0
Step 2: Gather up all the terms with ππ¦
ππ₯ on one side of the equation and isolate
ππ¦
ππ₯.
2π¦ ππ¦
ππ₯ + 2
ππ¦
ππ₯ = 4 β 2π₯
β (2π¦ + 2) ππ¦
ππ₯ = 4 β 2π₯
β ππ¦
ππ₯ =
4 β 2π₯
2π¦ + 2
Step 3: Find the slope of the tangent at the given point [find ππ¦
ππ₯ at (π₯, π¦) = (5, β5)]
ππ¦
ππ₯ at (π₯, π¦) = (5, β5) is:
4 β 2(5)
2(β5) + 2 =
β6
β8 =
3
4
Step 4: Find the equation of the tangent, given a point on it (π₯1 , π¦1) and its slope (π):
π¦ β π¦1 = π(π₯ β π₯1)
β π¦ β (β5) = 3
4(π₯ β 5)
β π¦ + 5 = 3
4(π₯ β 5)
β 4π¦ + 20 = 3π₯ β 15
β 3π₯ β 4π¦ β 35 = 0
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 23
POINTS TO REMEMBER
1) If π¦ = π(π₯), then ππ¦
ππ₯= πβ²(π₯).
ππ¦
ππ₯ or πβ²(π₯) has many names. It can be called the differential of π with
respect to π, the derivative of π or π(π), the derived function of π(π₯), the differential coefficient of π(π₯), the
slope of the tangent to the curve at (π₯, π¦) or (π₯, π(π₯)), the gradient. [Note: πβ²(2) is the value of πβ²(π₯) when π₯ = 2]
2) ππππβπ
[π(π) Β± π(π)] = ππππβπ
π(π) Β± ππππβπ
π(π)
3) ππππβπ
[π(π). π(π)] = ππππβπ
π(π) . ππππβπ
π(π)
4) ππππβπ
[π(π)
π(π)] =
ππππβπ
π(π)
ππππβπ
π(π)
5) 0
sinlim 1
β=
6) dy
dx =
0limhβ
π(π₯ + β) β π(π₯)
β where π¦ = π(π₯).
7) If π¦ = π(π₯) = π₯π, then ππ¦
ππ₯ or πβ²(π₯) = π π₯πβ1. (*)
8) π¦ = π π(π₯),ππ¦
ππ₯= π πβ²(π₯), where π is a constant. [Multiplying constants can be brought outside and left there.]
Example: If π¦ = 3π₯5,ππ¦
ππ₯= 3 πβ²(π₯5) = 3 (5π₯4) = 15π₯4
9) If π(π₯) = π’(π₯) + π£(π₯), then df du dv
dx dx dx= + .
Example: If π¦ = 3π₯9 + 5π₯2 , ππ¦
ππ₯= 27π₯8 + 10π₯.
10) If π(π₯) = π’(π₯)π£(π₯), then d f dv duu v
dx dx dx= + . (*)
Example: If π¦ = (5π₯ + 2)π₯3 , ππ¦
ππ₯= (5π₯ + 2)3π₯2 + (π₯3)(5) β
ππ¦
ππ₯ = 20π₯3 + 6π₯2.
11) If π(π₯) = π’(π₯)
π£(π₯), then
2
du dvv u
df dx dx
dx v
β
= . (*)
Example: If π(π₯) = π₯2β 5
7π₯ ,
ππ¦
ππ₯=
(7π₯)(2π₯) β (π₯2β5)(7)
(7π₯)2 β ππ¦
ππ₯=
7π₯2 + 35
49π₯2
12) If π¦ = [π(π₯)]π β ππ¦
ππ₯ = π [π(π₯)]πβ1 [πβ²(π₯)].
Example: If π¦ = (4π₯3 β 5π₯ + 1)7 ,ππ¦
ππ₯ = 7(4π₯3 β 5π₯ + 1)6 (12π₯2 β 5) = (84π₯2 β 35) (4π₯3 β 5π₯ + 1)6
13) If π¦ = π ππ π₯, ππ¦
ππ₯= πππ π₯. If π¦ = π ππ (π(π₯)),
ππ¦
ππ₯ = [πππ (π(π₯)] [πβ²(π₯)].
Example: If π¦ = π ππ 14π₯2 ,ππ¦
ππ₯= (πππ 14π₯2) (28π₯).
14) If π¦ = πππ π₯, ππ¦
ππ₯= βπ ππ π₯. If π¦ = πππ (π(π₯)) ,
ππ¦
ππ₯ = [βπ ππ(π(π₯)] [πβ²(π₯)].
Example: If π¦ = πππ 5π₯ ,ππ¦
ππ₯= (βπ ππ 5π₯) (5) or β5 π ππ 5π₯.
15) If π¦ = π‘ππ π₯, ππ¦
ππ₯= π ππ2 π₯. If π¦ = π‘ππ (π(π₯)) ,
ππ¦
ππ₯ = [π ππ2(π(π₯)] [πβ²(π₯)].
Example: If π¦ = π‘ππ 8π₯ ,ππ¦
ππ₯= (π ππ2 8π₯) (8) or 8 π ππ2 8π₯ .
16) If π¦ = π ππβ1 π₯,ππ¦
ππ₯=
1
β1 β π₯2 . πΌπ π¦ = π ππβ1 π(π₯),
ππ¦
ππ₯=
1
β1β[π(π₯)]2 Γ πβ²(π₯).
Example: If π¦ = π ππβ1 βπ₯ , ππ¦
ππ₯=
1
β1 β (βπ₯ )2
Γ 1
2 (π₯)
β1
2 = 1
β1βπ₯ Γ
1
2βπ₯ =
1
2βπ₯ β1βπ₯ =
1
2 βπ₯βπ₯2
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 24
17) If π¦ = π ππβ1 π₯
π,
ππ¦
ππ₯=
1
βπ2 β π₯2 .
Example: If π¦ = π ππβ1 5π₯3
4,
ππ¦
ππ₯=
1
β16 β (5π₯3)2 Γ 15π₯2
18) If π¦ = π‘ππβ1 π₯,ππ¦
ππ₯=
1
1 + π₯2 . πΌπ π¦ = π‘ππβ1 π(π₯),ππ¦
ππ₯=
1
1 + [π(π₯)]2 Γ πβ²(π₯).
Example: If π¦ = π‘ππβ1(7π₯ β 3) , ππ¦
ππ₯=
1
1 + (7π₯β3)2 Γ 7 = 7
1 + (7π₯β3)2 = 7
49π₯2β 42π₯ + 10
19) If π¦ = π‘ππβ1 π₯
π,
ππ¦
ππ₯=
π
π2+ π₯2 .
Example: If π¦ = π‘ππβ1 5π₯3
4,
ππ¦
ππ₯=
4
(16 + (5π₯3)2 Γ 15π₯2 or 60π₯2
16 +25π₯6 .
20) If π¦ = ππ₯ ,ππ¦
ππ₯= ππ₯. If π¦ = ππ(π₯) ,
ππ¦
ππ₯= ππ(π₯) Γ πβ²(π₯).
Example: If π¦ = πβ9π₯+5 , ππ¦
ππ₯= πβ9π₯+5 Γ β9 = β9 πβ9π₯+5
21) If = ππππ π₯ ,ππ¦
ππ₯=
1
π₯ . If π¦ = ππππ π(π₯) ,
ππ¦
ππ₯=
πβ²(π₯)
π(π₯) . [i.e. the differential of the function over the function]
Example: If π¦ = ππππ(2π₯5 β 6π₯ + 7), ππ¦
ππ₯=
10π₯4 β 6
2π₯5β 6π₯ + 7 .
22) If π¦ = π(π₯), then π2π¦
ππ₯2 = π
ππ₯ (
ππ¦
ππ₯ ) = πβ²β² (π₯).
Example: If π(π₯) = π₯2 β 4π₯ + 7 β πβ²(π₯) = 2π₯ β 4 β πβ²β²(π₯) = 2.
23) Before attempting to differentiate difficult log functions, check if the question be made a easier by applying the
rules of logs.
24) ππ¦
ππ₯ is the slope of a tangent to a curve at any point on the curve.
25) To find a turning point on a graph , use ππ¦
ππ₯ = 0 and solve the resulting equation.
26) At local maximum: ππ¦
ππ₯= 0 and
π2π¦
ππ₯2 < 0
27) At local minimum: ππ¦
ππ₯= 0 and
π2π¦
ππ₯2 > 0
28) At point of inflection: ππ¦
ππ₯= 0 and
π2π¦
ππ₯2 = 0
29) Curves are increasing when ππ¦
ππ₯> 0. Curves are decreasing when
ππ¦
ππ₯< 0.
30) In reciprocal graphs such as π = ππ+π
ππ+π , the vertical asymptote is got by letting the denominator = 0 β π =
βπ
π .
Horizontal asymptotes are got by letting π¦ = ππππ₯ββ
π(π₯) and dividing above and below by highest power of π₯.
Here, horizontal asymptote is: = ππππ₯ββ
π(π₯) = ππππ₯ββ
π+ π
π₯
π+ π
π₯
= π+π
π+π β π =
π
π .
31) If π = distance travelled or displacement, π£ = ππ
ππ‘ = speed or velocity and π =
ππ£
ππ‘ =
π2π
ππ‘2 = acceleration.
32) With questions on rates of change of one variable with respect to another, always write down the βgimmeβ,
(e.g. ππ
ππ‘ = 0.75) , then write down what you are trying to find (e.g.
ππ΄
ππ‘ @ π = 5), do the sideway kick
(e.g. ππ΄
ππ‘ =
ππ΄
ππΓ
ππ
ππ‘ , then express one variable in terms of the other (e.g. π΄ = π π2) and differentiate
(e.g. ππ΄
ππ= 2ππ).
Finally, put in the βatβ value given.
33) Familiarise yourself with formulae available on pages 25 and 27 of Mathematical Formulae and Tables.
CALCULUS NOTES
Β© H Dorgan IOE Maths HL 25