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Mathematics AA SL IB Diploma Program - Topic 5: Calculus
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Detailed notes for mathematics analysis and approaches SL topic 5: calculus. Includes worked examples and IB-style questions
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topics 5 :Calculus
, Monday ,
8 March
introduction ntoroalculus
⑤ '
'
Calculus '
Latin word -_ pebble .
⑤
Differential calculus
*
Derivatives .
'
undamental Theory of calculus
*
Differentiation .
⑤
Integral calculus
*
Integrals .
*
Integration .
⑤
History :
⑨
Sir Isaac Newton vs .
Wilhelm Leibniz .
⑨
Newton accused Leibniz of had not published
plagerism although he his
findings on calculus .
⑨
As president of the royal society he biased a court to attribute the credit to him .
⑨
Atone time in history ,
calculus -
-
math .
⑨
Calculus t
founded in concept of LIMITS .
⑤
Lim fix )=A " LIMIT
X sa
"
as X approaches a .
⑤
Finding the limit :
⑨ ⑤ ⑤
Graph :
Lim fix )=x2 -
X -
6 Equation :L .
Lim fix )=x3
-
2x -15 Table : lim
fcx )=2
-15=33-213
X→2 x→3
X
)
5=27-6+5
t
X flx)
=
271-5 4.9 1.9
LIMIT -
- 26 4.99 1.99
XZ -
I
lim 4.999 1.999
" fix)
ztromtherictht
> 2. = X -
I
[
it
Iffy
x -
ex th
approaches
-
= g. Gaga i. 9999
f- ( x ) : -4 "
LIMIT
'
-
Xt 1 as X approaches 5
, fix) = 2
✓ approaches 2-
2. from the left f)
= It 1
LIMIT -_ 2
,⑤ Basic derivative rule :
⑨ t
f' ( x ) nxn
" -
fix )=X
' -
-
⑨ "
f' ( x ) if prime of x ( Notation ) .
1.
Bring down exponent to
multiply .
2. Subtract 1 to exponent .
⑤
Examples :
I
fix) 3×4-5×2
-
- -
Sx 2/0=1 2
fix
)=}xs -
8×4-531×3 -125×2 tax
3
gcx ) : # # # -
t -
8x
f' ( x ) 12×3-10×-5 f' 1×1=31×4-32×3 355×2+5×+9 g. ( x ) = 5×-4 3×-3+2×-2 8x
-
- - -
-
'
( x) 20×-5+9×-4 4×-3 8
g.
= -
-
-
⑤ First principle of derivatives : gicx)=
-
+
IT ¥-8 -
⑨ fxth ) fix)
f' ( x ) -_ lim -
h- O h
>
Where does it come from ?
*
Derivative is the instantaneous rate of change .
TANGENT LINE : line that touches a cuneata point .
*
Instantaneous rate of change = Gradient of a
tangent line .
SECANT LINE : line that crosses a curve at 2 Points .
spate of change = Gradient of a secant line .
y
*
Gradient -
-
X
⑨
Example : ( ( x) -_ 200×2-5×+80
"
C' 1×1=400×-5 derivative of a constant -_
Zero .
V
Rate of change of Ctx )
⑨
Derivative "
also known as the gradient function .
⑨ When substitute values of X. then you get the gradient and the tangent line at that value of
you x .
Examples Find the derivative the first principles
using
: :
I 2
fcx)= 3×-5
fcx )
-
-
XZ -
2×+1
( im flxth ) f ( x) *
ftxth ) *
flex )= fcx )=3X 5
f' ( x ) him f- ( x )
-
=
fix)=x2 -2×+1
-
-
h
h- O h
* h- O to
fcxth) =3 ( Xth ) -
5
fcxth )
-
- (Xth )2 -
2(xthltl
Lim 3Cxth ) -
S -
( 3×-5 ) Iim X72Xhth2 -
2x -
2h -11 -
( XZ 2Xt1 )
-
=x2t2xhth2 2. htt
-
2x -
h- O h h
n- O
( im # + 3h -15-3/7 # Iim xztzxhth
'
- 2x -
2. htt -212+2×-1
h→O h h- O h
( im 3K Iim 2x h th -
2h
h→0 ht n→0 h
Lim 3 =
3 lim hl2xth -21 -
2x th -
2 ' 2×+0-2
h
h→0 into
=
2X -2
, ⑨
Types of discontinuity :
⑨
Essential .
⑤ Removable .
)§
qghifnq.at gtfo
'" "
.ME#ugae.f'
=3
x' Fat
'
fix F f- "' '
a
-
lim fix) DNE
Xt -4 does not exist
*
The only way to remove a
discontinuity is to assign a function value at the point of discontinuity .
⑤ Limits at
infinity :
2X -
3 lim
*
f- ( X )=2
f- (x ) Xt 't x - + as
-
-
a Lim
*
as
fcx )=2
-
x→
-
) x' I
Iim "
← *
fix )
7 fix) F
t too it
fix
-
-_ -
x→ y
y=2
- -
, - .
-
>
lim
*
It
x→
fix)= as DNE
- -
-
-
O l l l l l l t CS
-
-
-
21=-7
V
HOMEWORK :
*
Solve the basic derivative rule :
using
F- X1 3×3 Ix't 5×-8 Ex 2 5×2-531×+9 Ex Ex3:hlX)=5 # tf ¥3 -
fix) gtx)
: -
-
- : -
-
- -
f' ( x ) 9×2-83×+5
-
-
g. (x ) -
- 5×2 -
21+9-5×-1 hlx) -
- 5 Ix
-
7×-1+8×-2 -
- 3
'
( X) 10x 55+5×-2 n' 1×1=7×-2 16×-3+21×-4
g.
-
- -
-
'
(x) tox Bt # n' ( ) ¥2 ¥t¥
g. x
- -
- -
- -
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