Core Maths C3
Revision Notes
October 2012
Core Maths C3 1
Algebraic fractions .................................................................................................. 3 Cancelling common factors .................................................................................................................. 3 Multiplying and dividing fractions ....................................................................................................... 3 Adding and subtracting fractions .......................................................................................................... 3 Equations .............................................................................................................................................. 4
2
Functions................................................................................................................. 4 Notation ................................................................................................................................................ 4 Domain, range and graph ...................................................................................................................... 4 Defining functions ............................................................................................................................................... 5
Composite functions ............................................................................................................................. 6 Inverse functions and their graphs ........................................................................................................ 6 Modulus functions ................................................................................................................................ 7 Modulus functions y = |f (x)|.............................................................................................................................. 7 Modulus functions y = f (|x|)............................................................................................................................... 8
Standard graphs..................................................................................................................................... 8 Combinations of transformations of graphs .......................................................................................... 9
3
Trigonometry ......................................................................................................... 10 Sec, cosec and cot ............................................................................................................................... 10 Graphs ................................................................................................................................................................ 10
Inverse trigonometrical functions ....................................................................................................... 11 Graphs ................................................................................................................................................................ 11
Trigonometrical identities ................................................................................................................... 11 Finding exact values .......................................................................................................................................... 12 Proving identities. .............................................................................................................................................. 12 Eliminating a variable between two equations ................................................................................................. 12 Solving equations .............................................................................................................................................. 12
R cos(x + α) ........................................................................................................................................ 13
4
Exponentials and logarithms ................................................................................. 14 Natural logarithms .............................................................................................................................. 14 Definition and graph .......................................................................................................................................... 14 Equations of the form eax + b = p ....................................................................................................................... 15
5
Differentiation ........................................................................................................ 15 Chain rule ............................................................................................................................................ 15 Product rule ......................................................................................................................................... 16 Quotient rule ....................................................................................................................................... 16 Derivatives of ex and logex ≡ ln x. ................................................................................................... 17
dy 1 = dx ............................................................................................................................................ 18 dx dy Trigonometric differentiation.............................................................................................................. 18 Chain rule – further examples ............................................................................................................. 19 Trigonometry and the product and quotient rules ............................................................................... 19
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6
Numerical methods ................................................................................................ 20 Locating the roots of f(x) = 0 ............................................................................................................. 20 The iteration xn + 1 = g(xn) ................................................................................................................... 20 Conditions for convergence................................................................................................................. 21
Index ............................................................................................................................. 22
2
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Algebraic fractions
Cancelling common factors x 3 − 2 x 2 − 3x . 4 x 2 − 36
Example:
Simplify
Solution:
First factorise top and bottom fully –
x( x 2 − 2 x − 3) x( x − 3)( x + 1) x 3 − 2 x 2 − 3x = = 4( x − 3)( x + 3) 4( x 2 − 9) 4 x 2 − 36 and now cancel all common factors, in this case (x – 3) to give x( x + 1) . 4( x + 3)
Answer =
Multiplying and dividing fractions This is just like multiplying and dividing fractions with numbers and then cancelling common factors as above. 9 x 2 − 4 3x 2 − x − 2 ÷ 3x 2 − 2 x x3 + x2
Example:
Simplify
Solution:
First turn the second fraction upside down and multiply
=
9x 2 − 4 x3 + x2 × , and then factorise fully 3x 2 − 2 x 3x 2 − x − 2
=
(3 x − 2)(3x + 2) x 2 ( x + 1) × , and cancel all common factors (3 x + 2)( x − 1) x(3 x − 2) Answer
x( x + 1) . x −1
=
Adding and subtracting fractions Again this is like adding and subtracting fractions with numbers; but finding the Lowest Common Denominator can save a lot of trouble later. 3x 5 . − 2 x − 7 x + 12 x − 4 x + 3
Example:
Simplify
Solution:
First factorise the denominators
2
=
3x 5 − ; ( x − 3)( x − 4) ( x − 3)( x − 1)
=
3x( x − 1) 5( x − 4) − ( x − 3)( x − 4)( x − 1) ( x − 3)( x − 1)( x − 4)
=
3x 2 − 3 x − 5 x + 20 3x 2 − 8 x + 20 = which cannot be simplified further. ( x − 1)( x − 3)( x − 4) ( x − 1)( x − 3)( x − 4)
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we see that the L.C.D. is (x – 3)(x – 4)(x – 1)
3
Equations x x −1 1 − = x +1 x 2
Example:
Solve
Solution:
First multiply both sides by the Lowest Common Denominator which in this case is 2x(x + 1)
2
⇒
x × 2x – (x – 1) × 2(x + 1) = x(x + 1)
⇒
2x2 – 2x2 + 2 = x2 + x
⇒
x2 + x – 2 = 0,
⇒
x = –2 or 1
⇒
(x + 2)(x – 1) = 0
Functions
A function is an expression (often in terms of x) which has only one value for each value of x.
Notation y = x2 – 3x + 7,
f (x) = x2 – 3x + 7
and
f : x → x2 – 3x + 7
are all ways of writing the same function.
Domain, range and graph The domain is the set of values which x can take: this is sometimes specified in the definition and sometimes is evident from the function: e.g. √x can only take positive x or zero. The range is that part of the y–axis which is used. Example:
Find the range of the function f : x → 2x – 3 with domain x is a real number: –2 < x ≤ 3.
Solution: First sketch the graph for values of x between – 2 and 3 and we can see that we are only using the y–axis from –7 to 3, and so the range is
y
-6
-4
-2
2
4
y is a real number: –7 < y ≤ 3. -5
4
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6
x 8
Example:
Find the largest possible domain and the range for the function
f : x → x − 3 + 1.
Solution: First notice that we cannot have the square root of a negative number and so x – 3 cannot be negative ⇒ x–3≥0 ⇒
largest domain is x real: x ≥ 3. y
To find the range we first sketch the graph and we see that the graph will cover all of the y-axis from 1 upwards and so the range is
Example: f:x→ Solution:
y real: y ≥ 1.
2
4
6
8
x
Find the largest possible domain and the range for the function 2x . x +1 The only problem occurs when the denominator is 0, and so x cannot be –1.
Thus the largest domain is x real: x ≠ –1. y
To find the range we sketch the graph
5
and we see that y can take any value except 2, so the range is y real: y ≠ 2.
-8
-6
-4
-2
2
4
6
8
x 1
Defining functions
Some mappings can be made into functions by restricting the domain. Examples: 1) The mapping x → √x where x ∈ ℜ is not a function as √–9 is not defined, but if we restrict the domain to positive or zero real numbers then f : x → √ x where x ∈ ℜ, x ≥ 0 is a function. 1 where x ∈ ℜ is not a function as the image of x = 3 is not defined, x −3 1 where x ∈ ℜ, x ≠ 3 is a function. but f : x → x −3 2)
x→
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Composite functions To find the composite function fg we must do g first. Example:
f : x → 3x – 2 and
Solution:
Think of f and g as ‘rules’
f is
times by 3
g is
square
⇒
fg is
g : x → x2 + 1. Find fg and gf.
subtract 2
add 1
square
add 1
times by 3
subtract 2
giving (x2 + 1) × 3 – 2 = 3x2 + 1 ⇒
gf is
fg : x → 3x2 + 1
times by 3
or
fg(x) = 3x2 + 1.
subtract 2
square
add 1
giving (3x – 2)2 + 1 = 9x2 – 12x + 5 ⇒
gf : x → 9x2 – 12x + 5
or gf (x) = 9x2 – 12x + 5.
Inverse functions and their graphs The inverse of f is the ‘opposite’ of f: thus the inverse of ‘multiply by 3’ is ‘divide by 3’ and the inverse of ‘square’ is ‘square root’. The inverse of f is written as f –1: note that this does not mean ‘1 over f ’. The graph of y = f –1(x) is the reflection in y = x of the graph of y = f (x) To find the inverse of a function x → y (i)
interchange x and y
(ii)
find y in terms of x.
Example:
Find the inverse of f : x → 3x – 2.
Solution:
We have x → y = 3x – 2
(i)
interchanging x and y ⇒ x = 3y – 2
(ii)
solving for y ⇒ y =
⇒
x+2 f –1 : x → 3
x+2 3
y=x y 2
-4
-2
2
y = f –1(x) -2
y = f (x)
6
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x
Example:
Find the inverse of g : x →
Solution:
We have x → y =
x+3 . 2x − 5
x+3 2x − 5
(i)
interchanging x and y y+3 ⇒ x = 2y − 5
(ii)
solving for y
y=x
⇒ x(2y – 5) = y + 3 ⇒
2xy – 5x = y + 3
⇒
2xy – y = 5x + 3
⇒
y(2x – 1) = 5x + 3
⇒
5x + 3 y = 2x − 1
⇒
g –1 : x →
Note that f f –1 (x) = f
–1
y 5 y = g –1(x) y = g (x)
5
x 1
5x + 3 2x − 1
f (x) = x.
Modulus functions Modulus functions y = |f (x)|
|f (x)| is the ‘positive value of f (x)’, so to sketch the graph of y = |f (x)| first sketch the graph of y = f (x) and then reflect the part(s) below the x–axis to above the x–axis. Example:
Sketch the graph of y = |x2 – 3x|.
Solution:
First sketch the graph of y = x2 – 3x.
y 5
Then reflect the portion between x = 0 and x = 3 in the x–axis
to give
5
y 5
5
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x
x
7
Modulus functions y = f (|x|)
In this case f (–3) = f (3), f (–5) = f (5), f (–8.7) = f (8.7) etc. and so the graph on the left of the y–axis must be the same as the graph on the right of the y–axis, so to sketch the graph, first sketch the graph for positive values of x only, then reflect the graph sketched in the y–axis. Sketch the graph of y = |x|2 – 3|x|.
Example:
y 5
Solution: First sketch the graph of y = x2 – 3x for positive values of x.
only -5
5
x
y 5
Then reflect the graph sketched in the y–axis to complete the sketch. -5
5
Standard graphs y=x4
y = x3 2
3
1 -4
y=x2
y 4
y=x
3y
2
-2
2
x
4
1
-1
-4
-2
-3
-2
-1
1
2
3
4
x
1
-3
2
y = x and y = x4 y = x and y = x3
y 2
4y 3
1 2 -4
-3
-2
-1
1
2
3
4
x 1
-1 -2
-1
1
2
3
4
5
6
-1 3
1 y= x 8
y = √x C3 14/04/13 SDB
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x
x
Combinations of transformations of graphs We know the following transformations of graphs: y = f (x) ⎛ a⎞ translated through ⎜ ⎟ becomes y = f (x - a) + b ⎝ b⎠
stretched factor a in the y-direction becomes y = a × f (x) ⎛ x⎞ stretched factor a in the x-direction becomes y = f ⎜ ⎟ ⎝ a⎠
reflected in the x-axis becomes y = – f (x) reflected in the y-axis becomes y = f (–x) We can combine these transformations: Examples:
1)
y = 2f (x – 3) is the image of y = f (x) under a stretch in the y-axis of factor 2 ⎛ 3⎞ followed by a translation ⎜⎜ ⎟⎟ , or the translation followed by the stretch. ⎝0⎠
2)
y = 3x2 + 6 is the image of y = f (x) under a stretch in the y-axis of factor 3 ⎛ 0⎞ followed by a translation ⎜⎜ ⎟⎟ , ⎝ 6⎠ BUT these transformations cannot be done in the reverse order.
To do a translation before a stretch we have to notice that ⎛0⎞ 3x2 + 6 = 3(x2 + 2) which is the image of y = f (x) under a translation of ⎜⎜ ⎟⎟ ⎝ 2⎠ followed by a stretch in the y-axis of factor 3.
3)
y = – sin(x + π) is the image of y = f (x) under a reflection in the x-axis followed ⎛−π ⎞ ⎟⎟ , or the translation followed by the reflection. by a translation of ⎜⎜ ⎝ 0 ⎠
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3
Trigonometry
Sec, cosec and cot 1 ; cos x
Secant is written sec x =
cotangent is written cot x =
cosecant is written cosec x =
1 and sin x
1 cos x = tan x sin x
Graphs y
y
1
y 1
1
90
180
270
360
x
90
-1
180
270
360
x
90
180
270
360
x
-1 -1
y = sec x
y = cosec x
y = cot x
Notice that your calculator does not have sec, cosec and cot buttons so to solve equations involving sec, cosec and cot, change them into equations involving sin, cos and tan and then use your calculator as usual. Example:
Find cosec 35o.
Solution:
cosec 35o =
Example:
Solve sec x = 3.2
Solution:
sec x = 3.2
⇒
10
1 1 = 1.743 to 4 S.F. o = 0.53576... sin 35
⇒
for 0 ≤ x ≤ 2πc 1 1 = 3.2 ⇒ cos x = = 0.3125 cos x 3.2
x = 1.25 or 2π – 1.25 = 5.03 radians
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Inverse trigonometrical functions The inverse of sin x is written as arcsin x or sin–1 x and in order that there should only be one value of the function for one value of x we restrict the domain to –π/2 ≤ x ≤ π/2 . Similarly for the inverses of cos x and tan x, as shown below. Graphs y 3
y
y
1 1
2 -2
-1
1
2
x 1
-2
-1
1
2
-1 -2
-1
1
y = arcsin x
y = arccos x
–1 ≤ x ≤ 1
–1 ≤ x ≤ 1
–π/2 ≤ arcsin x ≤ π/2
0 ≤ arccos x ≤ π
-1
x
2
y = arctan x x∈ℜ π
– /2 < arctan x < π/2
Trigonometrical identities You should learn these sin2 θ + cos2 θ = 1 tan θ + 1 = sec θ 2
2
1 + cot2θ = cosec2θ
tan( A - B) = tan 2 A =
sin (A + B) = sin A cos B + cos A sin B sin (A – B) = sin A cos B – cos A sin B sin 2A = 2 sin A cos A cos (A + B) = cos A cos B – sin A sin B
tan 3A =
tan A - tan B 1 + tan A tan B
2 tan A 1 - tan 2 A 3 tan A − tan 3 A 1 − 3 tan 2 A
sin P + sin Q = 2 sin
P+Q P -Q cos 2 2
sin P – sin Q = 2 cos
P+Q P -Q sin 2 2
cos P + cos Q = 2 cos
P+Q P -Q cos 2 2
cos (A – B) = cos A cos B + sin A sin B cos 2A
= cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A
sin2 A = ½ (1 – cos 2A) cos2 A = ½ (1 + cos 2A) sin2 ½ θ cos2 ½ θ
= ½ (1 – cos θ) = ½ (1 + cos θ)
cos P – cos Q =
– 2 sin
P+Q P -Q sin 2 2
2 sin A cos B = sin(A + B) + sin(A – B)
sin 3A = 3 sin A – 4 sin3 A
2 cos A sin B = sin(A + B) – sin(A – B)
cos 3A = 4 cos3A – 3 cos A
2 cos A cos B = cos(A + B) + cos(A – B)
tan( A + B) =
tan A + tan B 1 - tan A tan B
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–2 sin A sin B = cos(A + B) – cos(A – B)
11
x
Finding exact values
When finding exact values you may not use calculators. Example:
Find the exact value of cos 15o
Solution:
We know the exact values of sin 45o, cos 45o and sin 30o, cos 30o
so we consider =
1 2
×
3 2
cos 15 = cos (45 – 30) = cos 45 cos30 + sin 45 sin 30 +
1 2
× 12 =
3 + 1
6 + 4
=
2 2
2
.
Example: Given that A is obtuse and that B is acute, and sin A = 3/5 and cos B = 5/13 find the exact value of sin (A + B). Solution: We know that sin (A + B) = sin A cos B + cos A sin B so we must first find cos A and sin B.
Using
sin2 θ + cos2 θ = 1
⇒
cos2 A = 1 – 9/25 =
⇒
cos A = ± /5 and sin B = ± /13
16
/25
and sin2 B = 1 – 25/169 =
4
144
/169
12
But A is obtuse so cos A is negative and B is acute so sin B is positive ⇒ ⇒
cos A = – 4/5 and sin B =
12
sin (A + B) = 3/5 × 5/13 + – 4/5 × 12/13 =
/13 –33
/65
Proving identities.
Start with one side, usually the L.H.S., and fiddle with it until it equals the other side. Do not fiddle with both sides at the same time. cos 2 A + 1 = cot 2 A . 1 - cos 2 A
Example:
Prove that
Solution:
cos 2 A + 1 2 cos 2 A - 1 + 1 2 cos 2 A = = L.H.S. = = cot 2 A . Q.E.D. 2 2 1 - cos 2 A 1 - (1 - 2sin A) 2 sin A
Eliminating a variable between two equations
Example:
Eliminate θ from x = sec θ – 1, y = tan θ.
Solution:
We remember that tan2θ + 1 = sec2θ
⇒
sec2θ – tan2θ = 1.
secθ = x + 1 and tanθ = y ⇒
(x + 1)2 – y2 = 1
⇒
y2 = (x + 1)2 – 1 = x2 + 2x.
Solving equations
Here you have to select the ‘best’ identity to help you solve the equation. Example:
Solve the equation sec2A = 3 – tan A, for 0 ≤ A ≤ 360o.
Solution:
We know that
⇒
12
tan2θ + 1 = sec2θ
tan2A + 1 = 3 – tan A C3 14/04/13 SDB
⇒
tan2A + tan A – 2 = 0,
⇒
(tan A – 1)(tan A + 2) = 0
⇒
tan A = 1 or tan A = –2
⇒
A = 45, 225, or 116.6, 296.6.
factorising gives
R cos(x + α) An alternative way of writing a cos x ± b sin x using one of the formulae listed below (I)
R cos (x + α) = R cos x cos α – R sin x sin α
(II)
R cos (x – α) = R cos x cos α + R sin x sin α
(III)
R sin (x + α) = R sin x cos α + R cos x sin α
(IV)
R sin (x – α) = R sin x cos α – R cos x sin α
To keep R positive and α acute we select the formula with corresponding + and – signs. The technique is the same which ever formula we choose. Example:
Solve the equation 12 sin x – 5 cos x = 6 for 0o ≤ x ≤ 3600.
Solution: First re–write in the above form: notice that the sin x is +ve and the cos x is –ve so we need formula (IV). R sin (x – α) = R sin x cos α – R cos x sin α = 12 sin x – 5 cos x
Equating coefficients of sin x, ⇒ R cos α = 12
[i]
Equating coefficients of cos x, ⇒ R sin α = 5
[ii]
Squaring and adding [i] and [ii] ⇒ R2 cos2 α + R2 sin2 α = 122 + 52 ⇒
R2 (cos2 α + sin2 α) = 144 + 25
⇒
R2 = 169
but cos2 α + sin2 α = 1
⇒ R = ±13
But choosing the correct formula means that R is positive ⇒ R = +13 Substitute in [i] ⇒ ⇒
cos α =
12
/13
α = 22.620o or ....
But choosing the correct formula means that α is acute ⇒
⇒
α = 22.620o.
12 sin x – 5 cos x = 13 sin(x – 22.620)
and so to solve
12 sin x – 5 cos x = 6
we need to solve
13 sin(x – 22.620) = 6
⇒
sin(x – 22.620) = 6/13
⇒
x – 22.620 = 27.486 or 180 – 27.486 = 152.514
⇒
x = 50.1o or 175.1o .
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Example: Find the maximum value of 12 sin x – 5 cos x and the value(s) of x for which it occurs. Solution:
From the above example 12 sin x – 5 cos x = 13 sin(x – 22.6).
The maximum value of sin(anything) is 1 and occurs when the angle is 90, 450, 810 etc. i.e. 90 + 360n ⇒
the max value of 13 sin(x – 22.6) is 13
when x – 22.6 = 90 + 360n ⇒ x = 112.6 + 360no.
4
Exponentials and logarithms
Natural logarithms y
y = ex
y=x
Definition and graph 2
e ≈ 2.7183 and logs to base e
y = ln x 4
2
2 -
4 -
log e x is usually written ln x.
6 x
are called natural logarithms. 2 -
Note that y = ex and y = ln x 4 -
are inverse functions and that the graph of one is the reflection of the other in the line y = x.
Graph of y = e(ax + b) + c.
y = e2x y = ex y 6
The graph of y = e2x is the graph of
4
y = ex stretched by a factor of 1/2 in
2
the direction of the x-axis.
x
2
2 -
14
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The graph of y = e(2x + 3) is the graph of ⎛− y = e2x translated through ⎜⎜ ⎝ 0
3 2
2x + 3 = 2(x –
−3 2
y = e2x + 3 + 4
⎞ ⎟⎟ since ⎠
y 6
4
)
2
and the graph of y = e(2x +3) + 4 is that of
x −8
⎛0⎞ y = e(2x + 3) translated through ⎜⎜ ⎟⎟ ⎝ 4⎠
−6
−4
−2
2
y = e2x + 3
Equations of the form eax + b = p
Example:
Solve e2x + 3 = 5.
Solution: Take the natural logarithm of each side, remembering that ln x is the inverse of ex.
5
⇒
ln(e2x + 3) = ln 5
⇒
x =
⇒
2x + 3 = ln 5
ln 5 − 3 = –0.695 to 3 S.F. 2
Example:
Solve ln(3x – 5) = 4.
Solution: ln x.
Raise both sides to the power of e, remembering that ex is the inverse of
⇒
eln(3x – 5) = e4 ⇒
⇒
x =
(3x – 5) = e4
e4 + 5 = 19.9 to 3 S.F. 3
Differentiation
Chain rule If y is a composite function like y = (5x2 – 7)9 think of y as y = u9, where u = 5x2 – 7 then the chain rule gives
⇒ ⇒
dy dy du = × dx du dx dy du = 9u 8 × dx dx dy = 9(5 x 2 − 7) 8 × (10 x) = 90 x(5 x 2 − 7) 8 . dx
The rule is very simple, just differentiate the function of u and multiply by
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du . dx
15
dy . dx
Example:
y = √(x3 – 2x). Find
Solution:
y = √(x3 – 2x) = ( x 3 − 2 x) 2 .
⇒ ⇒ ⇒ ⇒
1
Put u = x3 – 2x
1
y= u 2 dy dy du = × dx du dx −1 dy du = 12 u 2 × = dx dx dy 3x 2 − 2 . = 1 dx 2( x 3 − 2 x) 2
1 2
( x 3 − 2 x)
−1
2
× (3x 2 − 2)
Product rule If y is the product of two functions, u and v, then dy dv du y = uv ⇒ . = u + v dx dx dx Example:
Differentiate y = x2 × √(x – 5).
Solution:
y = x2 × √(x – 5) = x2 × ( x − 5)
so put u = x2 and v = ( x − 5) dy dv du ⇒ = u + v dx dx dx = x2 × =
1 2 2
( x − 5)
x
2 x−5
−1
2
1
1
2
2
1
+ ( x − 5) 2 × 2x
+ 2x x − 5 .
Quotient rule If y is the quotient of two functions, u and v, then du dv v − u dy u dx dx . ⇒ y = = v dx v2 Example:
Differentiate y =
Solution:
y =
⇒
16
2x − 3 x 2 + 5x
2x − 3 , so put u = 2x – 3 and v = x2 + 5x 2 x + 5x du dv v − u dy dx dx = 2 dx v C3 14/04/13 SDB
=
( x 2 + 5 x) × 2 − (2 x − 3) × (2 x + 5) ( x 2 + 5 x) 2
=
2 x 2 + 10 x − (4 x 2 + 4 x − 15) − 2 x 2 + 6 x + 15 . = ( x 2 + 5 x) 2 ( x 2 + 5 x) 2
Derivatives of ex and logex ≡ ln x. y = ex
⇒
dy = ex dx
y = ln x
⇒
dy 1 = dx x
y = ln kx
⇒
y = ln k + ln x
⇒
dy 1 1 = 0+ = dx x x
y = ln xk
⇒
y = k ln x
⇒
dy k 1 = k× = dx x x
or use the chain rule. Example:
Find the derivative of f (x) = x3 – 5ex at the point where x = 2.
Solution:
f (x) = x3 – 5ex
⇒
f ′(x) = 3x2 – 5ex
⇒
f ′(2) = 12 – 5e2 = –24.9 f (x) = ln 3x – ln x5
Example:
Differentiate the function
Solution:
f (x) = ln 3x – ln x5 = ln 3 + ln x – 5 ln x = ln 3 – 4 ln x
⇒
f ′(x) =
−4 x
or we can use the chain rule
⇒
f ′(x) =
1 1 −4 × 3 − 5 × 5x 4 = x 3x x
Example:
Find the derivative of
Solution:
f (x) = log 10 3x = =
⇒
ln 3 x ln 10
(using change of base formula)
ln 3 + ln x ln 3 ln x = + ln 10 ln 10 ln 10
f ′(x) = 0 +
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f (x) = log 10 3x.
1 x
ln 10
=
1 . x ln 10 17
dy . dx where u = x2 eu, dy du = × du dx du = eu × = eu × 2x dx 2
Example:
y = e x . Find
Solution:
y = dy dx dy dx
⇒ ⇒
= 2x × e x . 2
dy . dx Solution: y = ln u, where u = 7x3 dy dy du ⇒ = × dx du dx dy 1 du 1 3 = × = . ⇒ × 21x 2 = 3 dx u dx x 7x Example:
y = ln 7x3. Find
dy 1 = dx dx dy Example:
x = sin2 3y. Find
Solution:
First find
dy . dx
dx as this is easier. dy
dx = 2 sin 3 = 2 sin 3y cos 3y × 3 = 6 sin 3y cos 3y dy ⇒
1 1 dy 1 = = = dx = 6 sin 3 y cos 3 y 3 sin 6 y dx dy
1 3
cosec 6y
Trigonometric differentiation x must be in RADIANS when differentiating trigonometric functions.
18
f (x)
f ′ (x)
sin x
cos x
cos x
– sin x
tan x
sec2 x
sec x
sec x tan x
cot x
– cosec2 x
cosec x
– cosec x cot x
important formulae sin2 x + cos2 x = 1 tan2 x + 1 = sec2 x 1 + cot2 x = cosec2 x C3 14/04/13 SDB
Chain rule – further examples Example:
y = sin4 x.
Solution: ⇒
y = sin4 x. y = u4 dy dy du = × dx du dx dy du = 4u 3 × dx dx
⇒ ⇒
Find
dy . dx
Put u = sin x
=
4 sin 3 x × cos x .
dy . dx
Example:
y = e sin x . Find
Solution:
y = eu, where u = sin x dy dy du = × dx du dx dy du = esin x × cos x = e sin x × dx dx
⇒ ⇒
= cos x × esin x .
dy . dx
Example:
y = ln (sec x). Find
Solution:
y = ln u, where u = sec x dy dy du = × dx du dx dy 1 du 1 = = × × sec x tan x = tan x . dx u dx sec x
⇒ ⇒
Trigonometry and the product and quotient rules Example:
Differentiate
y = x2 × cosec 3x.
Solution: y = x2 × cosec 3x so put u = x2 and v = cosec 3x dy du dv ⇒ = u + v dx dx dx 2 = x × (– 3cosec 3x cot 3x) + cosec 3x × 2x = – 3x2 cosec 3x cot 3x + 2x cosec 3x.
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19
Example:
Differentiate y =
Solution:
y =
⇒ ⇒ ⇒
6
tan 2 x 7x3
tan 2 x , so put u = tan 2x and v = 7x3 7x3 7 x 3 × 2 sec 2 2 x − tan 2 x × 21x 2 dy = dx (7 x 3 ) 2 14 x 3 sec 2 2 x − 21x 2 tan 2 x dy = dx 49 x 6 2 2 x sec 2 x − 3 tan 2 x dy = dx 7x 4
Numerical methods
Locating the roots of f(x) = 0 A quick sketch of the graph of y = f (x) can help to give a rough idea of the roots of f (x) = 0. If y = f (x) changes sign between x = a and x = b and f (x) is continuous in this region then a root of f (x) = 0 lies between x = a and x = b. Example:
Show that a root of the equation f (x) = x3 – 8x – 7 = 0 lies between 3 and 4.
Solution:
f (3) = 27 – 24 – 7 = –4,
and f (4) = 64 – 32 – 7 = +25
Thus f (x) changes sign and f (x) is continuous ⇒ there is a root between 3 and 4.
The iteration xn + 1 = g(xn) Many equations of the form f (x) = 0 can be re–arranged as x = g (x) which leads to the iteration x n + 1 = g (x n). Example:
Show that the equation
can be re–arranged as x = Solution:
x3 – 8x – 7 = 0
⇒
3
x3 – 8x – 7 = 0
8 x + 7 and find a solution correct to four decimal places. x3 = 8x + 7
⇒
x=
3
8x + 7 .
From the previous example we know that there is a root between 3 and 4, so starting with x1 = 3 we use the iteration xn + 1 =
to give
3
8xn + 7
x1 = 3 x2 = 3.14138065239 x3 = 3.17912997899 x4 = 3.18905898325 x5 = 3.19166031127
y -4
-2
2
4
-5
-10
x6 = 3.19234114002 -15
x7 = 3.19251928098 x8 = 3.19256588883
-20
x9 = 3.19257808284 20
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x
As x8 and x9 are both equal to 3.1926 to 4 D.P. we conclude that a root of the equation is 3.1926 to 4 D.P. Note that we have only found one root: the sketch shows that there are more roots.
Conditions for convergence If an equation is rearranged as x = g(x) and if there is a root x = α then the iteration xn+1 = g(xn), starting with an approximation near x = α will converge if -1 < g ′(α) < 1
(i) y
y
x
x1
x2
x
x3
x2 x4
x3
x1
x5
(a) will converge without oscillating if 0 < g ′(α) < 1,
(ii)
will oscillate and converge if –1 < g ′(α) < 0,
y
will diverge if g ′(α) < -1 or
(b)
g ′(α) > 1.
10
x1 x2
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x
x3
21
Index algebraic fractions adding and subtracting, 3 equations, 4 multiplying and dividing, 3 chain rule, 15 further examples, 19 cosecant, 10 cotangent, 10 derivative ex, 17 ln x, 17 differentiation
dy 1 = dx dx dy
, 18
trigonometric functions, 18 equations graphical solutions, 20 exponential eax + b = p, 15 graph of y = e(ax + b) + c, 14 graph of y = ex, 14 functions, 4 combining functions, 6 defining functions, 5 domain, 4 inverse functions, 6 modulus functions, 7 range, 4
22
graphs combining transformations, 9 standard functions, 8 iteration conditions for convergence, 21 xn + 1 = g(xn), 20 logarithm graph of ln x, 14 natural logarithm, 14 product rule, 16 further examples, 19 quotient rule, 16 further examples, 19 R cos(x + α), 13 secant, 10 trigonometrical identities, 11 trigonometry finding exact values, 12 harder equations, 12 inverse functions, 11 proving identities, 12
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