Differentiation of Algebraic
Differentiation of Algebraic
Differentiation of Algebraic
Differentiation
Contents
1. DIFFERENTIATION OF ALGEBRAIC
1.1 Basic Differentiation (1st Principal Theory)………………………………….
1.2 Methods of Differentiation:
a. Basic Rules (Power Rule) ……………………………………………………….
b. Sum, Product & Quotient Rule ………………………………………………..
c. Chain Rule ……………………………………………………………………………..
1.3 Higher –order Derivatives ……………………………………………………………
2. DIFFERENTIATION OF EXPONENTIAL
3. DIFFERENTIATION OF LOGARITHMS
4. DIFFERENTIATION OF TRIGONOMETRY
4.1 Basic Differentiation of 6 trigonometry ……………………………………….
“Hexagon Wheel of Trigo” – differentiation …………………………….
4.2 Rules of Differentiation of Trigonometry
in o tan …………………………………..
e ot ……………………………………
……………………………………………………………………..
…………………………………………………………………………..
4.3 Differentiation of trigonometry involve:
Algebraic ……………………………………………………………………………….
Exponential …………………………………………………………………………...
Logarithm ……………………………………………………………………………..
5 IMPLICIT DIFFERENTIATION
6 PARAMETRIC DIFFERENTIATION
2 Differentiation
DIFFERENTIATION OF ALGEBRAIC
OVERVIEW
f ( x h) f ( x )
f '( x) lim
h0 h
st
1 Principal
f x k , k is constant
d n 1
f x n f x f ' x
n
f ' x 0 DIFFERENTIATION
dx
Formula
Methods dy dy du
Basic
Chain Rule
dx du dx
f x xn , n
f ' x nx x 1
Sum, product f x u x v x
Product,
Quotient
f ' x u ' v uv '
f x kx n , n , k constant
quotient
sum
f ' x nkx x 1
u x
f x
v x
f x u x v x
f ' x u ' v uv '
f ' x u ' x v ' x
3 Differentiation
Any gradient on the graph can be found by differentiate the function to obtain gradient function.
differentiate
f x f ' x
‘Gradient function’
1
m=2
0.5 m=+ve
-2 -1 2 4 x
x
The gradient is constant (same) at
The gradient is different at every point
every point on the graph
on the graph.
f (x)
To find the gradient at point P x, f x ,
f ( x h) Q take another point on graph, let say point
Q with coordinate x h, f x h .
Q1
*h is increment between 2 points.
Q2 The gradient between point P and Q can
Q3 be found but not accurate. Now, take
P another point which is more closer to P
f (x) N that is Q1,Q2 and Q3. As h become smaller
(h approaching 0, h0) , the gradient will
x
x x+h more accurate.
f ( x h) f ( x )
f '( x) lim
h0 h
h
4 Differentiation
EXAMPLE 1
Solution:
f x 2x2 1
f x h 2 x h 1
2
f ( x h) f ( x) 2 x h 1 2 x 1
2
2
f '( x)
h h
2 x 2 2hx h 2 1 2 x 2 1
h
2 x 4hx 2h 2 2 x 2
2
h
h 4 x 2h
h
f '( x) lim 4 x 2h 4 x
h 0
# f ' x 4x is gradient function. The gradient will be different when x 1, x 5 . Let us gradient
method; m y yo .
x xo
#Gradient at x 1 , y 3 , take another point let say;
19 3 : m 3.205 3 4.1
x1 3 , y1 19 : m 8 x4 1.05, y4 3.205
3 1 1.05 1
x2 2 , y2 9 : m 93 6 x4 1.00001, y4 3.00004 : m
3.00004 3
4
2 1 1.00001 1
3.88 3
x3 1.2, y3 3.88 : m 4.4
1.2 1
*As x approaches to 1, the gradient gives constant value. Try using f ' x 4x when x 1 ;
f ' 1 4 1 4
TRY
The f ' x when x 5 .
5 Differentiation
Basic Rules
BASIC
Exp 2:
Exp 3:
f x abc f ' x 0 1. f x x3 f ' x 3x2
f x 1001 f ' x 0 2. 2 2 13 2 1 1 1
f x f x x f ' x x 3
f x ln101 f ' x 0 33 x 3 3 3
f x log 202 f ' x 0 Change from ‘surd form’ 2 4 2
x 3
f x e111 f ' x 0
to ‘exponential form’
9 9 3 x4
3. f t 2t 3 f ' t 2 3 t 4
*since f x don’t have variable
x, so the f ' x 0
Exp 4:
2
f x 4 x3 3 x x ln 3 e3
3x
Change form:
i) Surd exponential
1
ii) Division multiplication 2 1
f ' x 4 x3 3x 2 x x ln 3 e3
3
1 2 2
1
ln 3 and e3 constants
4 3 x 3 x 1 x 2 ln 3
2
2 3
3 2
12 x 2 ln 3
2 x 3x 2
6 Differentiation
x3
a. f x 8x 2 4 x 7 ; x 3 b. f x 3x ; x6
2
4
c. f x 6 x3 ; x 1 d. f x 11x2 x ; x 2
x2
2
e. f x x2 4 x 5 ; x0 f. f x 4x ; x4
x
dy
3. Find the .
dx
f x 25x2
2
a. b. f x 13x3 3x 3
x2 4
f x
6
c. f x x d.
2 x 3x 2
5
e. f x x ln 4 f. f x x5 7 x3 e
3
3
g. f x ln 4 x 3
9 x2 2 x h. f x e2 7ln 3
4x
3
c. Identify which is a constant. Then differentiate.
7 Differentiation
f x u x v x u x
f x u x v x f x
v x
f ' x u ' x v ' x
f ' x u ' v uv ' u ' v uv '
f ' x
v2
REFER TO Exp 4
Apply when numerator has x
EXAMPLE 5 Sum rule variable, and denominator
also has x variable and has
Find the f ' x for f x x2 x 1 more than 2 terms.
Solution:
f x x2 x 1 f x x3 x2
Differentiate by terms: f ' x 3x2 2x
Find the f ' x for f x x 3 2 x 1
2
Solution:
f x x 2 3 2 x 1
2 x3 x 2 6 x 3
f ' x 2 3 x 2 2 x 6
6x2 2x 6
8 Differentiation
Idea of thinking: (1) It can be expanded since the there has power of 1.
or;(2) Can use product rule
Solution: v
u
f x x5 2 x3 1
;Since it is product function so we take x as u and 2 x 1 as v.
5 3
u x5 v 2 x3 1
u ' 5x4 v ' 6 x2
Find the f ' x for f x x 2 2 x x
2
2
Solution:
u x2 2 v 2x2 x
u ' 2x v ' 4x 1
f ' x u ' v uv '
2 x 2 x 2 x x 2 2 4 x 1
4 x3 2 x 2 4 x3 x 2 8x 2
8 x3 3x 2 8 x 2
9 Differentiation
2 3
Find the f ' x for f x 2 x
x x
2
Solution:
2 x2 7
Find the f ' x for f x
x
Solution:
- The numerator and denominator both have x variable so quotient rule can be use but
there is easy way since the denominator only have 1 terms and numerator has power of 1.
- Separate the fraction into 2 fractions.
f x 2 x 7 x 1
- Write in 1 line 4 x x 1 2 x 2 7
- differentiate Subs into f ' x
x2
f ' x 2 7x 2 formula
& 2x2 7
7 simplify;
2 x2
x2
7
2 2
x
10 Differentiation
4 x 3
3
Idea of thinking: there is x variable at numerator (above part) & denominator (below part)
Solution:
12 4 x 3 x 3 4 x x 3 4 x 3
2 2 3
2 2
# tips: remember quotient rule
f x Common mistake:
x 3 2
2 2
Forget the order of u’v - uv’.
Who come first?
4 4 x 3 x 3 3 x 3 4 x 3
2 2 2
1st : write uv – uv
x 3
4
2
2nd : put prime (‘) at first letter
and last letter that is u and v.
4 4 x 3 3 x 2 9 4 x 3
2
x 3
3
2
11 Differentiation
1
There is composite function. Exp: x2 7 , 2 x 3 , x 11 2 .
3 5
Use when
dy dy du
If y f u and u g x then, . It called chain rule.
dx du dx
Alternatively, using formula: if y f x ,then;
n
dx
Chain rule normally compatible with “LET U method” (my students always say that). Chain rule is
very important in ‘Application of Differentiation’ where the rate of changes problem occurs.
4
Find the dy for y 3 x 5
3
dx
9 x
dy 4 1
Subs into formula: 4 3x3 5 2
dy dy du dx
3
dx du dx 36 x 2 3 x 3 5
4u 3 9 x 2
36 x 2u 3
3
36 x 2 3x3 5
Other example where can use chain rule (by definition @ formula), recommended to use formula for
time saving.
1.
f x x 4 5x 3.
4 3 4
x f x
2 2 3
3x 5
2.
3x 4.
4
3
5 f x
3
f x 2
x 3
2
3x3 5 3
**Example (1) & (2): Has to use product rule and quotient rule. By taking u and v specifically,
differentiate them using chain rule (by formula).
**Example (3): Change the surd form to exponential form, then multiply with power (i.e. 4). To use
chain rule, the function must be written as y f x .
n
1
**Example (4): Review n
a n . So it becomes, f x 3 3x3 5 3
a
13 Differentiation
2
Find the dy for y
dx 1 2x
Solution: ‘Inner ‘function as u
Can written as; 1 Using ;
2 2
y
d
f x n f x f ' x
n n 1
1 2x dx
2
2 1 2 x
1
u 1
1 2x y u2
du
2 11 2 x 2
11
dy 1 12
dx u 1
du 2 2 2
1
4 1 2 x
2
2 1 2 x
1 2
y
1 2x
Subs into formula: 1
2 1 2 x
2
dy dy du
dx du dx 1 1
2 1 2 x 2 2
dy 1
1 1 2
u 2 4 1 2 x
2 dx
2 3
2 1 2 x
1 2
1 2 x
2
2u 2
1
2 2 1 2 x 1 2 x
1 2 2
1 1
2 2 1 2 x 2 1 2 x
2
2
3
2 1 2 x
2
2
2 2
3
2 1 2 x
2
The best way to differentiate is by letting the coefficient ‘stay’ in front of the function we want to
1
1
differentiate to avoid messy and mistake. Exp: 2 1 2x 2 1 2x
1 2
2
Coefficient
Differentiate this part only
14 Differentiation
2
2 3
Find the d y2 for y 8 x 3 x2 5
2
dx x
Solution:
1)find dy
dx
1
3
y 8x2 2x 2
3 x2 5
1 1
1
2x
dy 31
8 2 x 21 2 x 2 3 3 x 2 5
dx 2
3
2
16 x x 2
18 x x 2 5
2
2)then find d y2
dx
3 2
; 18 x x 2 5
dy 2
16 x x 2 18 x x 2 5 has to solved by product rule
dx
2
since there 2 function here; 18x and x 2 5 .
Use product rule:
u 18 x
2
d
v x2 5 18 x x 2 5 18 x 4 x x 2 5 18 x 2 5
2 2
u ' 18
dx
v ' 2 x2 5 2x
18 x 2 5 4 x x 2 5
4x x 5 2
18 x 2
5 x 4 x 5
2
2
Then for d y2 ;
dx
2
3 52
d y
dx 2
16
2
x 18 x 2 5 x 2 4 x 5
15 Differentiation
Solution:
f ' x 25 x 4 2 x 2
f '' x 100 x 3 4 x 3
f ''' x 300 x 2 12 x 4
f 4 x 600 x 48 x 5
ANSWERS: example 1
1 a) 6 x 6 d) 9 x2 14 x 5
b) 3x2 6 x e) 2x 6
c) 2x 7 f) 20x3
2 a) 52 d) 43
b) 57 e) 4
c) 26 5
f)
8
3 a) 100x3 e) ln 4
1
25 x 4 21 x
b) 39 x 2 2 x 3 f)
3 2
c) x 1 6 x 2 9x 2
g) ln 4 18 x 2
4
d) 8 3
x h) 0
3