Numbers system)‫(نظام االعداد‬
1) Natural numbers)‫(االعداد الطبيعية‬:
Are numbers in the form 𝑁 = {1,2,3, … . } which are closed)‫(مغلقة‬
under addition)‫ (الجمع‬and multiplication)‫(الضرب‬
2) Integer numbers)‫(االعداد الصحيحة‬:
Are numbers in the form 𝑍 = {0, ±1, ±2, ±3, … . } which are
closed under addition and multiplication
3) Rational numbers)‫(االعداد القياسية‬:
𝑝
Are numbers in the form 𝑄 = {𝑞 ∶ 𝑝, 𝑞 ∈ 𝑍 , 𝑞 ≠ 0} which
are closed under addition, subtraction)‫ (الطرح‬and multiplication
4) Irrational numbers)‫(االعداد غير القياسية‬:
Are numbers which are not rational 𝑄∗ = {𝑥 ∶ 𝑥 ∉ 𝑄} ,(i.e
𝑝
they can not expressed as 𝑞 ,such as √2 , √3 , √7 , 𝜋 , … )
5) Real numbers)‫(االعداد الصحيحة‬:
The set of rational and irrational numbers are called real numbers
𝑅 = 𝑄⋃𝑄∗ which are closed under addition, subtraction ,
multiplication and division(quotient))‫(القسمة‬
‫نصف العلم اخطر من الجهل‬
A little learning is a dangerous
1
Intervals)‫(الفترات‬
A subset )‫(المجموعة الجزئية‬of real line is called an interval )‫(فتره‬if it
contains at least two numbers and contains all real numbers lying
between any two of its elements.
Suppose that 𝑎, 𝑏 ∈ 𝑅 𝑎𝑛𝑑 𝑎 < 𝑏 then we define the:
1) Closed interval)‫ (فتره مغلقه‬as [𝑎, 𝑏] = {𝑥 ∈ 𝑅 ∶ 𝑎 ≤ 𝑥 ≤ 𝑏}
2) Open interval)‫ (فتره مفتوحه‬as (𝑎, 𝑏) = {𝑥 ∈ 𝑅 ∶ 𝑎 < 𝑥 < 𝑏}
3) Half Open )‫(نصف مفتوحه‬interval(half closed)‫ )(نصف مغلقه‬as
[𝑎, 𝑏) = {𝑥 ∈ 𝑅 ∶ 𝑎 ≤ 𝑥 < 𝑏}
4) Half Open interval(half closed) as (𝑎, 𝑏] = {𝑥 ∈ 𝑅 ∶ 𝑎 <
𝑥 ≤ 𝑏}
Inequalities)‫(المتباينات‬
The process)‫ (العملية‬of finding the interval or intervals of numbers
that satisfy equality in x is called solving)‫ (حل‬the equality.
Example:
Solve the following inequalities and show their solutions sets on
the real line:
(1) 2x − 1 < 𝑥 + 3
(4) − 2𝑥 + 3 ≤ 7
(2 )
−x
< 2𝑥 + 1
3
(5) 5 < 3𝑥 + 10 ≤ 16
2
(3 )
6
≥5
x−1
(6)12x − 18 > 0
Solutions:
(1) 2x − 1 < 𝑥 + 3 ⟹ 2𝑥 < x + 4 ⇒ x < 4
The solution set is open interval (−∞, 4)
(2 )
−x
< 2𝑥 + 1 ⟹ −𝑥 < 6𝑥 + 3 ⟹ 0 < 7𝑥 + 3
3
⟹ −3 < 7𝑥 ⟹
−3
<𝑥
7
3
The solution set is open interval (− 7 , ∞)
(3) The inequality
otherwise)‫(ماعدا ذلك‬
6
x−1
6
≥5
x−1
can hold only if x > 1, because
is undefined )‫(غير معرفه‬or negative.
Therefore x − 1 is positive)‫ (موجب‬and the inequality will be
preserved if we multiply both side by x − 1, and we have:
6
11
≥ 5 ⟹ 6 ≥ 5x − 5 ⟹ 11 ≥ 5x ⟹
≥x
x−1
5
11
The solution set is open interval (1, 7 )
(4) − 2𝑥 + 3 ≤ 7 ⟹ −2x < 4 ⟹ 𝑥 > −2
The solution set is open interval (−2, ∞)
(5) 5 < 3𝑥 + 10 ≤ 16
⟹
⟹ −5 < 3𝑥 ≤ 6
−5
<𝑥≤2
3
−5
The solution set is open interval ( 3 , 2]
3
(1)12x − 18 > 0 ⟹ 12𝑥 > 18
⟹𝑥>
3
2
3
The solution set is open interval (2 , ∞)
Example:
Solve the following inequality and show their solutions sets on the
real line:
(1) (x − 2)(x − 3) ≥ 0
(2) x 2 < 3𝑥 + 10
Solutions:
We have two cases)‫(لديناحالتان‬:
(i)
(x − 2) ≥ 0 and (x − 3) ≥ 0
⟹ x ≥ 2 and x ≥ 3 ⟹ [2, ∞) ∩ [3, ∞)
The solution from intersection)‫ (تقاطع‬is [3, ∞)
(ii)
(x − 2) ≤ 0 and (x − 3) ≤ 0
⟹ x ≤ 2 and x ≤ 3 ⟹ (−∞, 2] ∩ (−∞, 3]
The solution from intersection is (−∞, 2]
Then the solution of inequality is (𝑖 ) ∪ (ii)
I.e. (−∞, 2] ∪ [3, ∞)
----------------------------------------------------------------------------------------(2) x 2 < 3𝑥 + 10 ⟹ x 2 − 3x − 10 > 0
⟹ (x − 5)(x + 2) > 0
4
We have two cases:
Case one:
(X − 5) > 0 𝑎𝑛𝑑 (x + 2) < 0
(X − 5) > 0 ⟹ 𝑥 > 5 ⟹ (5, ∞)
And (x + 2) < 0 ⟹ 𝑥 < −2 ⟹ (−∞, −2)
The solution is (5, ∞) ∩ (−∞, −2) = ∅
Case two:
(X − 5) < 0 𝑎𝑛𝑑 (x + 2) > 0
(X − 5) < 0 ⟹ 𝑥 < 5 ⟹ (−∞, 5)
And (x + 2) > 0 ⟹ 𝑥 > −2 ⟹ (−2, ∞)
The solution is (−∞, 5) ∩ (−2, ∞) = (−2,5)
Then the solution of inequality is (𝑖 ) ∪ (ii) = (−2,5)
Example:
Solve the following inequality and show their solutions sets on the
(1 )
real line:
x
x−4
<0
(2 )
2𝑥+1
𝑥+3
>3
Solution:
(1 )
x
<0
x−4
Case(i):
x < 0 and (x − 1) > 0 ⟹ 𝑥 < 0 ⟹ (−∞, 0)
5
𝑎𝑛𝑑 𝑥 > 1 ⟹ (1, ∞)
⟹ (−∞, 0) ∩ (1, ∞)
The solution from intersection is ∅
Case(ii):
x > 0 and (x − 1) < 0
⟹ x > 0 ⟹ (0, ∞) and x < 1 ⟹ (−∞, 1)
⟹ (0, ∞) ∩ (−∞, 1)
The solution from intersection is (0,1)
Finally)‫ (اخيرا‬the solution of inequality is 𝑐𝑎𝑠𝑒(𝑖 ) ∪ case(ii)
(0,1) ∪ ∅ = (0,1)
⟹ x ∈ (0,1)
----------------------------------------------------------------------------------------(2 )
2𝑥 + 1
>3
𝑥+3
Case(i):
𝑥 + 3 > 0 ⟹ 2𝑥 + 1 > 3𝑥 + 9 ⟹ −8 > 𝑥
But 𝑥 + 3 > 0 ⟹ x > −3 so we reject)‫ (نرفض‬case one
Case(ii):
𝑥 + 3 < 0 ⟹ 2𝑥 + 1 < 3𝑥 + 9 ⟹ −8 < 𝑥
Where 𝑥 + 3 < 0 ⟹ x < −3
⟹ −8 < 𝑥 < −3
6
the solution of inequality is (−8, −3)
Example:
Solve the following inequality and show their solutions sets on the
(1)(x + 2)2 (x − 3) > 0 (2) (x − 1)2 (x + 3)3 ≤ 0
real line:
Solution:
(1)(x + 2)2 (x − 3) > 0
(x + 2)2 is always)‫ (دائما‬positive ⟹ (x − 3) > 0 ⟹ x > 3
So the solution of inequality is
(3, ∞)
----------------------------------------------------------------------------------------(2 ) (x − 1 )2 (x + 3 )3 ≤ 0
(x − 1)2 is always positive ⟹ (x + 3) ≤ 0
⟹ x ≤ −3 ⟹ (−∞, −3]
So the solution of inequality is (−∞, −3] ∪ {1}
Example:
Solve the following inequality and show their solutions sets on the
real line: 3 ≤ 2x + 1 ≤ 7
Solution:
3 ≤ 2x + 1 ≤ 7 ⟹ 2 ≤ 2x ≤ 6 ⟹ 1 ≤ x ≤ 3
⟹ x ∈ [1,3]
7
Example:
Solve the following inequality and show their solutions sets on the
real line:
2x + 3 ≤ 4x − 1 ≤ x + 11
Solution:
Firstly)‫ (اوال‬we take: 2x + 3 ≤ 4x − 1 ⟹ 4 ≤ 2x ⟹ 2 ≤ x
Secondly )‫(ثانيا‬we take: 4x − 1 ≤ x + 11 ⟹ 3x ≤ 12 ⟹ x ≤ 6
So x is satisfy)‫ (تحقق‬the inequality 2 ≤ x ≤ 6
⟹ [2,6]
Absolute value)‫(القيمه المطلقه‬
The absolute value of 𝑥 ∈ 𝑅 is defined as)‫(تعرف ب‬:
−𝑥,
|𝑥 | = {
𝑥,
𝑥<0
𝑥≥0
Some properties )‫(خواص‬of Absolute value:
(i)
(ii)
|−𝑎| = 𝑎
|𝑎 ± 𝑏 | = |𝑏 ± 𝑎 |
(iii)
|𝑏| = |𝑏|
(iv)
(v)
(vi)
|𝑎 ∙ 𝑏 | = |𝑎 | ∙ |𝑏 |
|𝑎 + 𝑏 | ≤ |𝑎 | + |𝑏 |
|𝑎 + 𝑏 | ≥ |𝑎 | − |𝑏 |
𝑎
|𝑎|
Representing )‫(تمثيل‬the absolute value in intervals
(1) 𝐼𝑓 |𝑥 | < 𝑎 ⟹ −a < 𝑥 < a
8
(2) 𝐼𝑓 |𝑥 | > 𝑎 ⟹ x > 𝑎 𝑜𝑟 𝑥 < −a
(3) 𝐼𝑓 |𝑥 | = 𝑎 ⟹ x ± a
Example:
Solve the following inequality:
(1 ) | 𝑥 − 5 | ≤ 3
(2) |2𝑥 + 1| > 6
(4)|3𝑥 − 7| = 8
2
(5) | − 3| < 5
𝑥
(3) |2𝑥 − 3| ≥ 1
Solution:
(1) |𝑥 − 5| ≤ 3 ⟹ −3 ≤ x − 5 ≤ 3 ⟹ 2 ≤ x ≤ 8
(2) |2𝑥 + 1| > 6 ⟹ −6 > 2𝑥 + 1 > 6 ⟹ −7 > 2𝑥 > 5
∴
−7
5
>𝑥>
2
2
(3) |2𝑥 − 3| ≥ 1 ⟹ −1 ≥ 2x − 3 ≥ 1 ⟹ 2 ≥ 2x ≥ −2
∴ 1 > 𝑥 > −1
(4)|3𝑥 − 7| = 8 ⟹ 3x − 7 = ±8
⟹ 3x − 7 = 8 and 3x − 7 = −8
−1
⟹ 3x − 7 = 8 ⟹ x = 5 and 3x − 7 = −8 ⟹ x =
3
So the solution is x = 5 and
2
(5) | − 3| < 5
𝑥
⟹ −5 <
x=
−1
3
2
2
− 3 < 5 ⟹ −2 < < 8
𝑥
𝑥
9
We have two cases:
Case one: 𝑥 > 0 ⟹ −2x < 2 < 𝑥8 ⟹ −x < 1 < 4𝑥
x>
1
4
𝑎𝑛𝑑 𝑥 < −1
case two : 𝑥 < 0 ⟹ −2x > 2 > 𝑥8 ⟹ −x > 1 > 4𝑥
x<
1
4
𝑎𝑛𝑑 𝑥 > −1
1
so the solution is two intervals (4 , ∞) 𝑎𝑛𝑑 (−∞, −1)
Exercise(1):
(i)Solve)‫ (حل‬the following inequalities and show their solutions sets
on the real line:
(1) 3x − 1 > 10
(2) 3 − 5𝑥 ≤ 17
(3)2𝑥 + 1 < 0
(4)(𝑥 − 3)(𝑥 + 1) ≥ 0
(5)(2𝑥 − 1)(𝑥 − 4) < 0
(6) 2𝑥 − 3 < 6𝑥 + 4 ≤ 𝑥 − 7
(7) 8 > 5𝑥 − 3 ≤ 12
(8 ) 𝑥 2 < 9
(9)
3𝑥 + 1
≥6
𝑥−2
(10) 𝑥 3 + 𝑥 2 ≥ 0
(11) (𝑥 − 3)(𝑥 − 2)(𝑥 + 1) ≥ 0
(13) 1 < 5 − 3𝑥 < 11
(15) (𝑥 − 2)(𝑥 − 6) > 0
(12) 3𝑥 − 4 ≤ 8
(14) (𝑥 + 1)2 (𝑥 + 3) > 0
(16) 𝑥(𝑥 − 2)(𝑥 + 3) > 0
(17) (𝑥 − 3)(𝑥 + 5)(𝑥 − 4)2 < 0 (18) 18𝑥 − 3𝑥 2 > 0
(19) (𝑥 − 1)2 (𝑥 + 4) > 0
(20) (3𝑥 − 1)(2𝑥 + 3) > 0
10
(21) (𝑥 − 4)(2𝑥 − 3) < 0
(22) (𝑥 − 2)3 > 0
(ii)Solve the following inequalities
(1)|3x − 1| > 1
(2 ) | 𝑥 + 2 | ≥ 8
2 − 3𝑥
|≤4
1 + 2𝑥
(4) |3𝑥 + 5| = 1
(5 ) |
(7 ) | 𝑥 + 2 | < 3
(8 ) | 𝑥 − 2 | ≥ 1
(3) |2𝑥 − 1| < 13
(6 ) |
2𝑥 − 5
|<3
𝑥−6
(9 ) | 𝑥 − 3 | < 1
𝑥
(10) |2𝑥 − 3| < 5 (11) | − 2| < 4 (12) |4𝑥 − 1| ≥ 1
3
1
3
4
(13) |2 + | > 1 (14) | − 2| ≤ 4 (15) | | < 3
𝑥
𝑥
𝑥
(16)|7𝑥 − 3| ≥ 1 (17)|3𝑥 + 7| < 2 (18)|𝑥 − 2| <
‘‫إن القرين الى المقارن ينسب‬
1
2
‫اختر قرينك واصطفيه تفاخرا‬
A man is known by the company he keeps
11
Functions)‫(الدوال‬
Let A and B be sets. A function from A to B is a rule for assigning to
each element in A one element in B. If 𝑓(𝑥 ) = 𝑦 then y is called
the value of x. x is called the independent)‫ (مستقل‬variable and y is
called the dependent )‫(تابع‬variable)‫(متغير‬.
Example:
If 𝑓(𝑥 ) = 2𝑥 2 − 4𝑥 + 3 𝑎𝑛𝑑 𝑔(𝑥 ) = 𝑥 2 + 5 . Find
(1) 𝑓(0 ) , 𝑓(−2), 𝑓(3) (2)𝑔(−4), 𝑔(𝑓(1))
Solution:
(1 ) 𝑓 (0 ) = 0 − 0 + 3 = 3
𝑓(−2) = 8 + 8 + 3 = 19
𝑓(3) = 18 − 12 + 3 = 9
(2) 𝑔(−4) = 16 + 5 = 21
𝑔(𝑓(1)) ⟹ f(1) = 1 ⟹ g(f(1)) = 1 + 5 = 6
Exercise:
x−1
1
(1)If 𝑓(𝑥 ) = x+1 prove)‫ (اثبت‬that 𝑓 (𝑥 ) = −𝑓(𝑥)
(2)If 𝑓(𝑥 ) = 𝑥 2
, 𝑔(𝑥 ) = 𝑥 3 find:
1
1
𝑔(𝑏)−𝑔(𝑎)
(i) 𝑓(−3) + 𝑓(−3) (ii) 𝑔 (6 ) − 𝑓 (6) (iii) 𝑓(𝑏)−𝑓(𝑎)
(3)If 𝑓(𝑥 ) = 3 prove that 𝑓(𝑥 + 1 ) = 9𝑓(𝑥 − 1)
5
(4)If 𝑓(𝑥 ) = 𝑥 prove that
(5)If 𝑓(𝑥 )√𝑥 prove that:
12
𝑓(𝑥+ℎ )−𝑓(𝑥)
ℎ
𝑓(𝑥+ℎ )−𝑓(𝑥)
ℎ
−5
= 𝑥(𝑥+ℎ)
=
1
√𝑥+ℎ+√𝑥
Type of functions)‫(انواع الدوال‬
(1)Polynomials)‫ (كثيرات الحدود‬functions:
A function of the form 𝑓(𝑥 ) = 𝑎0 𝑥 𝑛 + 𝑎1 𝑥 𝑛−1 + 𝑎2 𝑥 𝑛−2 + ⋯ + 𝑎𝑛
is called a polynomial function, where 𝑎0 + 𝑎1 + 𝑎2 + ⋯ + 𝑎𝑛 are
constants)‫ (ثوابت‬and n is natural )‫ (طبيعي‬number called the degree
)‫ (الدرجه‬of the polynomial (the domain of polynomial is R)
Examples:
(1)
(2)
𝑦 = 2𝑥 5 + 7𝑥 3 − 𝑥 2 + 9𝑥 + 1 is a polynomial of
five degree
𝑦 = 𝑥 2 + 9𝑥 is a polynomial of second degree
(2)Rational functions)‫(الدوال الكسرية‬:
Are the functions in the form
𝑃(𝑥)
𝑄(𝑥)
Where (𝑥) ≠ 0 , 𝑝(𝑥 )𝑎𝑛𝑑 𝑄(𝑥)
are polynomials functions
(3)Algebraic functions)‫(الدوال الجبرية‬:
Algebraic functions are relation)‫ (عالقه‬between x and y
satisfies)‫ (تحقق‬an equation)‫(المعادلة‬
𝑝0 𝑦 𝑛 + 𝑝1 𝑦 𝑛−1 + 𝑝2 𝑦 𝑛−2 + ⋯ + 𝑝𝑛 = 0 Where 𝑝0 + 𝑝1 + 𝑝2 +
⋯ + 𝑝𝑛 are polynomials.
(4)Exponential functions)‫(الدوال االسية‬:
Are the functions in the form 𝑓(𝑥 ) = 𝑎𝑢(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑎 > 1 the
number is called the base)‫ (االساس‬of the exponential function
13
3
Examples 42𝑥 , 7−𝑥 , 5𝑥
2 +9𝑥
, 𝑥 𝑠𝑖𝑛𝑥
Note )‫(ملحوظة‬:
(1) 𝑎 𝑥 ∙ 𝑎 𝑦 = 𝑎 𝑥+𝑦
𝑎𝑥
(2) 𝑦 = 𝑎 𝑥−𝑦
𝑎
(3 ) 𝑎 𝑥 = 𝑎 𝑦 ⟹ x = y
(4) (ax )y = axy
𝑎 𝑥 𝑎𝑥
(5 ) ( ) = 𝑥
𝑏
𝑏
(6 )
1
= 𝑎−𝑥
𝑥
𝑎
(5)Logarithmic functions)‫(الدوال اللوغرثمية‬:
Are the functions in the form 𝑓(𝑥 ) = 𝑙𝑜𝑔𝑎 ℎ(𝑥 ) 𝑤ℎ𝑒𝑟𝑒 𝑎 >
1 𝑖𝑓 𝑎 = 𝑒 then 𝑓(𝑥 ) = 𝑙𝑜𝑔𝑒 ℎ(𝑥 ) = 𝑙𝑜𝑔𝑒 ℎ(𝑥 ) = 𝑙𝑛ℎ(𝑥) is called
the natural )‫(الطبيعي‬logarithmic.
Note )‫(ملحوظه‬:
(1) 𝑙𝑜𝑔𝑎 (𝑥𝑦) = 𝑙𝑜𝑔𝑎 𝑥 + 𝑙𝑜𝑔𝑎 𝑦
𝑥
(2) 𝑙𝑜𝑔𝑎 ( ) = 𝑙𝑜𝑔𝑎 𝑥 − 𝑙𝑜𝑔𝑎 𝑦
𝑦
(3) 𝑙𝑜𝑔𝑎 𝑥 𝑛 = 𝑛𝑙𝑜𝑔𝑎 𝑥
(4) 𝑙𝑜𝑔𝑎 𝑎 = 1
(5) 𝑙𝑜𝑔𝑎 0 = −∞
14
(6) 𝑙𝑛𝑒 = 1
(7) 𝑒 𝑙𝑛𝑥 = 𝑥 ∀ 𝑥 > 0
(6)Trigonometric functions)‫(الدوال المثلثية‬:
Are the functions given in the form
𝑠𝑖𝑛𝑥
,
𝑐𝑜𝑠𝑒𝑐𝑥 =
𝑐𝑜𝑠𝑥
1
𝑠𝑖𝑛𝑥
,
𝑡𝑎𝑛𝑥 =
, 𝑐𝑜𝑡𝑥 =
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
, 𝑠𝑒𝑐𝑥 =
1
𝑐𝑜𝑠𝑥
1
𝑐𝑜𝑠𝑥
=
𝑡𝑎𝑛𝑥 𝑠𝑖𝑛𝑥
(7)Inverse)‫ (العكسية‬trigonometric functions:
Are the functions given in the form
arcsin 𝑥 𝑜𝑟 𝑠𝑖𝑛−1 𝑥 , 𝑎𝑟𝑐𝑐𝑜𝑠𝑥 𝑜𝑟 𝑐𝑜𝑠 −1 𝑥 , 𝑎𝑟𝑐𝑡𝑎𝑛𝑥 𝑜𝑟 𝑡𝑎𝑛−1 𝑥
𝑎𝑟𝑐𝑠𝑒𝑐𝑥 𝑜𝑟 𝑠𝑒𝑐 −1 𝑥, 𝑎𝑟𝑐𝑐𝑜𝑠𝑒𝑐𝑥 𝑜𝑟 𝑐𝑜𝑠𝑒𝑐 −1 𝑥, 𝑎𝑟𝑐𝑐𝑜𝑡𝑥 𝑜𝑟 𝑐𝑜𝑡 −1 𝑥
Properties)‫(خواص‬:
(1) 𝐼𝑓 𝑦 = 𝑠𝑖𝑛−1 𝑥 ⟹ 𝑥 = 𝑠𝑖𝑛𝑦
(2) 𝑠𝑖𝑛−1 𝑥 ≠ (𝑠𝑖𝑛𝑥 )−1 ≠
1
(𝑠𝑖𝑛𝑥 )
(3) 𝑠𝑖𝑛−1 (𝑐𝑜𝑠𝑥) = −𝑥 𝑎𝑛𝑑 𝑠𝑖𝑛−1 (𝑠𝑖𝑛𝑥 ) = 𝑥
This properties for all inverse trigonometric functions
(8)Hyperbolic functions)‫(الدوال الزائدية‬:
Are the functions given by exponential functions in the form
15
𝑒 𝑥 − 𝑒 −𝑥
𝑠𝑖𝑛ℎ𝑥 =
2
𝑒 𝑥 + 𝑒 −𝑥
, 𝑐𝑜𝑠ℎ𝑥 =
2
𝑠𝑖𝑛ℎ𝑥 𝑒 𝑥 − 𝑒 −𝑥
1
1
𝑡𝑎𝑛ℎ𝑥 =
= 𝑥
,
𝑠𝑒𝑐ℎ𝑥
=
,
𝑐𝑜𝑠𝑒𝑐ℎ𝑥
=
𝑐𝑜𝑠ℎ𝑥 𝑒 + 𝑒 −𝑥
𝑐𝑜𝑠ℎ𝑥
𝑠𝑖𝑛ℎ𝑥
𝑐𝑜𝑡ℎ𝑥 =
𝑐𝑜𝑠ℎ𝑥
𝑠𝑖𝑛ℎ𝑥
(9)Inverse Hyperbolic functions:
Are functions in the form
𝑠𝑖𝑛ℎ−1 𝑥 , 𝑐𝑜𝑠ℎ−1 𝑥, 𝑡𝑎𝑛ℎ−1 𝑥 , 𝑠𝑒𝑐ℎ−1 𝑥 , 𝑐𝑜𝑠𝑒𝑐ℎ−1 𝑥 , 𝑐𝑜𝑡ℎ−1 𝑥
The relation between inverse hyperbolic functions and
logarithmic functions:
−1
If 𝑠𝑖𝑛ℎ 𝑥 = 𝑦 ⟹ x = sinhy =
𝑒 𝑦 −𝑒 −𝑦
2
⟹ 𝑒 𝑦 − 𝑒 −𝑦 = 2𝑥 ⟹ (ey )2 − 1 = 2xey
⟹ (ey )2 − 2xey − 1 = 0
2x ± √4x 2 + 4
⟹e =
2
y
⟹ y = ln [x ± √x 2 + 1]
∴ 𝑠𝑖𝑛ℎ−1 x = ln [x ± √x 2 + 1]
16
(10)Even )‫(زوجيه‬and odd)‫ (فرديه‬functions:
The function 𝑓(x) is called
(1)Even if 𝑓(−x) = 𝑓(x)
(2)Odd if 𝑓(−x) = −𝑓(x)
Example:
Determine)‫ (حدد‬the even and odd function if:
(1)𝑓(x) = x 2 + 5 (2)𝑓(𝑥 ) = 𝑥 3 − 𝑥 (3)𝑓(𝑥 ) = 𝑠𝑖𝑛𝑥
(4 )𝑓 (𝑥 ) =
2
(5)𝑓(𝑥 ) = 3𝑥 4 + 7𝑥 − 1
2
𝑥 −4
Solutions:
(1)𝑓(−x) = (−x)2 + 5 = x 2 + 5 = f(x)
Since)‫(𝑓 (بما ان‬−x) = 𝑓(x) Then)‫ (اذا‬it is even function
______________________________________________________
(2)𝑓(−𝑥 ) = (−𝑥 )3 − (−𝑥 ) = −𝑥 3 + 𝑥 = −(𝑥 3 − 𝑥 ) = −𝑓(𝑥)
Since 𝑓(−x) = −𝑓(x) Then it is odd function
_______________________________________________________
(3)𝑓(−𝑥 ) = sin(−𝑥)
⟹ sin(0 − 𝑥 ) = 𝑠𝑖𝑛0𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠0𝑠𝑖𝑛𝑥 = −𝑠𝑖𝑛𝑥 = −𝑓(𝑥)
𝑓(−x) = −𝑓(x) Then it is odd function
17
_______________________________________________________
(4)𝑓(−𝑥 ) =
2
(−𝑥)2 −4
2
= 𝑥 2−4 = 𝑓(𝑥)
Since 𝑓(−x) = 𝑓(x) Then it is even function
_______________________________________________________
(5)𝑓(−𝑥 ) = 3(−𝑥)4 + 7(−𝑥) − 1 = 3𝑥 4 − 7𝑥 − 1
Since 𝑓(−𝑥 ) ≠ 𝑓(𝑥 ) ≠ −𝑓(𝑥) Then it is not even and not odd
Domain of definition )‫(مجال التعريف‬:
Let X and Y be sets and let 𝑓 be a function from X to Y. the set X is
called the domain of the function. Then the domain of definition of:
(1)Polynomials functions:
The domain of definition is 𝐷𝑓 = 𝑅 = (−∞, ∞)
Example:
Find the domain of definition
(1) 𝑓(𝑥 ) = 𝑥 2 + 3𝑥 − 20 (2) 𝑓(𝑥 ) = 3𝑥 4 − 𝑥 3 + 2𝑥 + 1
Solution:
(1) 𝑓(𝑥 ) = 𝑥 2 + 3𝑥 − 20
The domain of definition is: 𝐷𝑓 = 𝑅
(2) 𝑓(𝑥 ) = 3𝑥 4 − 𝑥 3 + 2𝑥 + 1
The domain of definition is: 𝐷𝑓 = 𝑅
18
(2)fraction)‫ (الكسريه‬functions:
Is the function in the form: 𝑓 =
𝑃(𝑥)
𝑄(𝑥)
Is defined in all 𝑅 except at points where 𝑄(𝑥) ≠ 0
𝐷𝑓 = 𝑅 − {𝑄(𝑥 ) = 0}
Example:
Find the domain of definition
(1 ) 𝑓 (𝑥 ) =
(4 ) 𝑓 (𝑥 ) =
3𝑥
𝑥−7
(2 ) 𝑓 (𝑥 ) =
2
𝑥
(3 ) 𝑓 (𝑥 ) =
2𝑥
𝑥 2 − 16
Solution:
3𝑥
⟹x−7≠0 ⟹x≠7
𝑥−7
(1 ) 𝑓 (𝑥 ) =
The domain of definition is:
𝐷𝑓 = R − {7}
(2 ) 𝑓 (𝑥 ) =
(3 ) 𝑓 (𝑥 ) =
2
⟹ x ≠ 0 ⟹ D = R − {0}
𝑥
3
√𝑥 − 5
⟹x−5>0 ⟹𝑥 >5
The domain of definition is:
𝐷𝑓 = (5, ∞)
19
3
√𝑥 − 5
(4 ) 𝑓 (𝑥 ) =
2𝑥
⟹ 𝑥 2 − 16 ≠ 0 ⟹ (x − 4)(x + 4) ≠ 0
2
𝑥 − 16
⟹ x ≠ 4 and x ≠ −4
The domain of definition is:
𝐷𝑓 = R − {4, −4}
(3)Roots )‫ (الجزرية‬functions:
The function in the 𝑓 = √𝑃(𝑥)
The domain of definition is:
𝐷𝑓 = 𝑃(𝑥 ) ≥ 0
Example:
Find the domain of definition
(1) 𝑓(𝑥 ) = √𝑥 2 − 9
(2) 𝑓(𝑥 ) = √4 − 𝑥 2
Solution:
(1) 𝑓(𝑥 ) = √𝑥 2 − 9 ⟹ x 2 − 9 ≥ 0 ⟹ 𝑥 2 ≥ 9
⟹ 𝑥 ≥ 3 𝑎𝑛𝑑 𝑥 ≥ −3
The domain of definition is
𝐷𝑓 = −3 ≥ 𝑥 ≥ 3
(2) 𝑓(𝑥 ) = √4 − 𝑥 2 ⟹ 4 − x 2 ≥ 0 ⟹ 4 ≥ 𝑥 2 ⟹ 𝑥 2 ≤ 4
⟹ x ≤ 2 and x ≤ −2
20
The domain of definition is
𝐷𝑓 = −2 ≤ 𝑥 ≤ 2
𝑜𝑟 𝐷 = [−2,2]
(4)logarithmic functions:
The function in the 𝑓(𝑥) = 𝑙𝑛𝑃(𝑥)
The domain of definition is:
𝐷𝑓 = 𝑃(𝑥 ) > 0
Example:
Find the domain of definition if :
𝑓(𝑥 ) = ln(𝑥 − 3)
Solution:
𝑓(𝑥 ) = ln(𝑥 − 3) ⟹ x − 3 > 0 ⟹ 𝑥 > 3
The domain of definition is:
𝐷𝑓 = (3, ∞)
Exercise:
Find the domain of definition if :
(A)
(1) 𝑓(𝑥 ) = 1 + √2𝑥
(2 ) 𝑓 (𝑥 ) =
(4) 𝑓(𝑥 ) = ln(𝑥 + 8) (5) 𝑓(𝑥 ) =
21
𝑥 2 −1
𝑥+5
−3
√𝑥
(3 ) 𝑓 (𝑥 ) =
2
𝑥 2 −2𝑥
(6) 𝑓(𝑥 ) = √|3𝑥 − 1|
‫‪(8)𝑓(𝑥 ) = √𝑥 2 − 64 (9)𝑓(𝑥 ) = √8 − 𝑥 2‬‬
‫‪1‬‬
‫𝑥‪1 − √2‬‬
‫= ) 𝑥( 𝑓 ) ‪(7‬‬
‫)‪(B‬‬
‫𝑥‪4‬‬
‫‪𝑥+5‬‬
‫= ) 𝑥( 𝑓 ) ‪(2‬‬
‫‪√𝑥 + 3‬‬
‫‪𝑥+3‬‬
‫= ) 𝑥( 𝑓 ) ‪(4‬‬
‫‪(1) 𝑓(𝑥 ) = 𝑥 3 − 6𝑥 2 + 9‬‬
‫𝑥‪2‬‬
‫‪𝑥 2 − 5𝑥 + 6‬‬
‫‪(6) 𝑓(𝑥 ) = √2𝑥 − 8‬‬
‫‪−5‬‬
‫‪√3𝑥 − 1‬‬
‫كلما ادبني الدهــــــــر‬
‫اراني ضعف عقلي‬
‫واذا ما ازددت علمــا‬
‫زادني علما بجهلي‬
‫‪Never too old to learn‬‬
‫‪22‬‬
‫= ) 𝑥( 𝑓 ) ‪(3‬‬
‫= ) 𝑥( 𝑓 ) ‪(5‬‬
Limits)‫(النهايات‬
Let a denote to )‫(تشير الى‬areal number and let 𝑓(𝑥 ) denote a
function, we say that L is the limit of 𝑓(𝑥 ) as x approaches to ‫(تقترب‬
)‫من‬a in this case, write
lim 𝑓(𝑥) = 𝐿
𝑥→𝑎
Example:
Determine
lim 2𝑥
𝑥→2
Let us make a table of values of x approaching 2 and the
corresponding)‫ (المطابقة‬value of 2x
x
2x
x
2x
2.1
4.2
1.9
3.8
2.01
4.02
1.99
3.98
2.001
4.002
1.999
3.998
.
.
.
.
.
.
.
.
.
.
.
.
2
4
2
4
x>2
x<2
As x approaches 2, we see that 2x approach 4, then lim 2𝑥 = 4
𝑥→2
Theorem)‫(نظريه‬:
Suppose )‫(افترض‬that lim 𝑓(𝑥) = 𝐴 𝑎𝑛𝑑 lim 𝑔(𝑥) = 𝐵 then:
𝑥→𝑎
(1 )
𝑥→𝑎
lim 𝑘 = 𝑘
𝑥→𝑎
23
(2 )
lim [𝑓(𝑥 ) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥) = 𝐴 ± 𝐵
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
(3) lim[𝑓(𝑥 ) ∙ 𝑔(𝑥)] = lim 𝑓(𝑥) ∙ lim 𝑔(𝑥) = 𝐴 ∙ 𝐵
𝑥→𝑎
𝑥→𝑎
𝑥→𝑎
(4) lim [𝑓(𝑥 ) ÷ 𝑔(𝑥)] = lim 𝑓(𝑥) ÷ lim 𝑔(𝑥) =
𝑥→𝑎
𝑥→𝑎
(5 )
𝑥→𝑎
𝑛
lim [𝑓(𝑥 )]𝑛 = [lim 𝑓(𝑥)] = 𝐴𝑛 𝑛 ∈ 𝑁
𝑥→𝑎
𝑥→𝑎
𝑛
(6) lim 𝑛√𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐴 𝐴 ≥ 0
𝑥→𝑎
𝑥→𝑎
(7 )
lim[𝑙𝑛(𝑓(𝑥))] = 𝑙𝑛 (lim 𝑓(𝑥))
𝑥→𝑎
𝑥→𝑎
Example:
Calculate )‫(احسب‬the following limits:
(1) lim 7
𝑥→2
(2) lim 𝑥 + 3
𝑥→5
(3) lim 𝑥 2 − 4𝑥
𝑥→−1
Solution:
(1) lim 7 = 7
𝑥→2
(2) lim 𝑥 + 3 = 5 + 3 = 8
𝑥→5
(3) lim 𝑥 2 − 4𝑥 = (−1)2 − 4(−1) = 5
𝑥→−1
Example:
Calculate the following limits:
24
A
𝑤ℎ𝑒𝑟𝑒 𝐵 ≠ 0
B
2𝑥 − 1
(1) lim
𝑥→∞ 𝑥 + 4
(3𝑥 + 1)(𝑥 − 7)
𝑥 2 + 2𝑥 + 3
(2) lim
(3) lim
𝑥→∞ 2𝑥 2 − 5
𝑥→∞ (𝑥 + 9)(𝑥 − 2)
Solution:
2𝑥 − 1 ∞
=
𝑥→∞ 𝑥 + 4
∞
(1) lim
2𝑥 1
1
−
2
−
2𝑥 − 1
𝑥=2
⟹ lim
= lim 𝑥 𝑥 = lim
4
4
𝑥→∞ 𝑥 + 4
𝑥→∞ 𝑥
𝑥→∞
+
1
+
𝑥 𝑥
𝑥
𝑥 2 + 2𝑥 + 3 ∞
(2) lim
=
𝑥→∞ 2𝑥 2 − 5
∞
𝑥 2 2𝑥 3
2 3
+
+
1
+
2
2
2
𝑥 + 𝑥2 1
𝑥
𝑥
𝑥
⟹ lim
= lim
=
5
𝑥→∞ 2𝑥 2
𝑥→∞
5
2
2− 2
− 2
2
𝑥
𝑥
𝑥
(3𝑥 + 1)(𝑥 − 7) ∞
=
𝑥→∞ (𝑥 + 9)(𝑥 − 2)
∞
(3) lim
3𝑥 1 𝑥 7
( 𝑥 + 𝑥 ) (𝑥 − 𝑥 )
1
7
(3 + 𝑥 ) (1 − 𝑥 )
(3)(1)
⟹ lim
= lim
=
=3
9 𝑥 2
9
2
𝑥→∞ 𝑥
𝑥→∞
(1)(1)
( + )( − )
(1 + ) (1 − )
𝑥 𝑥 𝑥 𝑥
𝑥
𝑥
Example:
Calculate the following limits:
𝑥2 − 1
(1) lim
𝑥→1 𝑥 − 1
𝑥 2 − 5𝑥 + 6
𝑥3 + 8
(2) lim
(3) lim
𝑥→3
𝑥→−2 𝑥 + 2
𝑥−3
25
(4) lim
𝑥−1
𝑥→1 √𝑥
−1
Solution:
𝑥2 − 1 0
(1) lim
=
𝑥→1 𝑥 − 1
0
(𝑥 − 1)(𝑥 + 1)
𝑥2 − 1
⟹ lim
= lim
= lim 𝑥 + 1 = 2
𝑥→1 𝑥 − 1
𝑥→1
𝑥→1
(𝑥 − 1 )
𝑥 2 − 5𝑥 + 6 0
(2) lim
=
𝑥→3
𝑥−3
0
(𝑥 − 3)(𝑥 − 2)
𝑥 2 − 5𝑥 + 6
⟹ lim
= lim
= lim(𝑥 − 2) = 1
𝑥→3
𝑥→3
𝑥→3
(𝑥 − 3 )
𝑥−3
𝑥3 + 8 0
(3) lim
=
𝑥→−2 𝑥 + 2
0
(𝑥 + 2)(𝑥 2 − 2𝑥 + 4)
𝑥3 + 8
⟹ lim
= lim
𝑥→−2 𝑥 + 2
𝑥→−2
(𝑥 + 2 )
= lim (𝑥 2 − 2𝑥 + 4) = 4 + 4 + 4 = 12
𝑥→−2
(4) lim
𝑥−1
𝑥→1 √𝑥
⟹ lim
−1
𝑥−1
𝑥→1 √𝑥
−1
=
0
0
= lim
(𝑥 − 1)(√𝑥 + 1)
𝑥→1 (√𝑥
− 1)(√𝑥 + 1)
= lim (√𝑥 + 1) = 2
𝑥→1
26
(𝑥 − 1)(√𝑥 + 1)
𝑥→1
𝑥−1
= lim
Important limits)‫(النهايات الهامه‬:
𝑠𝑖𝑛𝑥
=1
𝑥→0 𝑥
(1) (𝑖) lim
𝑥
=1
𝑥→0 𝑠𝑖𝑛𝑥
(𝑖𝑖 ) lim
𝑎 𝑥
(2) (𝑖 ) (1) lim [1 + ] = 𝑒 𝑎
𝑥→∞
𝑥
(𝑖𝑖) lim[1 +
𝑥→0
1
𝑎𝑥 ]𝑥
= 𝑒𝑎
𝑥 𝑚 − 𝑎𝑚 𝑚 𝑚−𝑛
(3) lim 𝑛
= [𝑎 ]
𝑥→𝑎 𝑥 − 𝑎𝑛
𝑛
Example:
Calculate the following limits:
𝑠𝑖𝑛2𝑥
(1) lim
𝑥→0
𝑥
𝑥2
(2) lim
𝑥→0 𝑠𝑖𝑛2 7𝑥
𝑠𝑖𝑛3𝑥
𝑥→0 𝑠𝑖𝑛8𝑥
(3) lim
Solution:
𝑠𝑖𝑛2𝑥
𝑠𝑖𝑛2𝑥
= 2lim
= 2 (1 ) = 2
𝑥→0
𝑥→0 2𝑥
𝑥
(1) lim
𝑥2
𝑥
𝑥
(2) lim
=
lim
(
)
(
)
𝑥→0 𝑠𝑖𝑛2 7𝑥
𝑥→0 𝑠𝑖𝑛7𝑥
𝑠𝑖𝑛7𝑥
1
7𝑥
1
7𝑥
1
= ( lim
) ( lim
)=
7
𝑠𝑖𝑛7𝑥
7 𝑥→0 𝑠𝑖𝑛7𝑥
49
𝑥→0
𝑠𝑖𝑛3𝑥
𝑠𝑖𝑛3𝑥 𝑥
𝑠𝑖𝑛3𝑥
𝑥
= lim (
) ( ) = lim (
) lim (
)
𝑥→0 𝑠𝑖𝑛8𝑥
𝑥→0 𝑠𝑖𝑛8𝑥
𝑥→0
𝑥→0 𝑠𝑖𝑛8𝑥
𝑥
𝑥
(3)lim
𝑠𝑖𝑛3𝑥 1
8𝑥
1
3
= 3lim (
) ( ) lim (
) = 3( ) =
𝑥→0
3𝑥
8
𝑠𝑖𝑛8𝑥
8
8
𝑥→0
27
Example:
Find the following limits:
3 𝑥
6 𝑥+8
1 2𝑥−5
(1) lim [1 + ] (2) lim [1 − ]
(3) lim [1 + ]
𝑥→∞
𝑥→∞
𝑥→∞
𝑥
𝑥
𝑥
1
(4) lim[1 + 9𝑥 ]𝑥
𝑥→0
Solution:
3 𝑥
(1) lim [1 + ] = 𝑒 3
𝑥→∞
𝑥
6 𝑥+8
6 𝑥
68
(2) lim [1 − ]
= lim [1 − ] [1 − ]
𝑥→∞
𝑥→∞
𝑥
𝑥
𝑥
6 𝑥
68
6 𝑥
= lim [1 − ] lim [1 − ] = lim [1 − ] = 𝑒 −6
𝑥→∞
𝑥→∞
𝑥 𝑥→∞
𝑥
𝑥
1 2𝑥−5
1 2𝑥
1 −5
(3) lim [1 + ]
= lim [1 + ] lim [1 + ]
𝑥→∞
𝑥→∞
𝑥
𝑥 𝑥→∞
𝑥
2
2
1 𝑥
1 −5
1 𝑥
[ lim [1 + ] ] [1 + ] = [ lim [1 + ] ] = [𝑒]2 = 𝑒 2
𝑥→∞
𝑥→∞
𝑥
∞
𝑥
(4)lim[1 +
𝑥→0
1
7𝑥 ]𝑥
= 𝑒7
28
Exercise:
Find the following limits
(A)
√𝑥 2 + 2
(1)lim
𝑥→0 3𝑥 − 6
(2)lim8
(3) (4)lim(−3)
𝑥→3
𝑥→0
(𝑥 2 + 12𝑥 3 − 17𝑥 + 2)
(5) lim
𝑥→1
𝑥+3
(4) lim 2𝑥
𝑥→−4
𝑥−2
𝑦+6
6𝑥 − 9
(
)
(
)
6 lim 2
7 lim 3
𝑥→∞ 𝑥 2 + 2𝑥 + 1
𝑦→6 𝑦 − 36
𝑥→0 𝑥 − 12𝑥 + 3
(5) lim
𝑡3 + 8
𝑥 2 − 4𝑥 + 4
(8) lim
(9)lim 2
𝑡→−2 𝑡 + 2
𝑥→2 𝑥 + 𝑥 − 6
𝑥 2 − 16
(11)lim
𝑥→4 𝑥 − 4
3𝑥 − 3−𝑥
(13)lim 𝑥
𝑥→0 3 + 3−𝑥
√𝑥 − 2
(12)lim 2
𝑥→2 𝑥 − 4
3𝑥 − 3−𝑥
(13) lim 𝑥
𝑥→−∞ 3 + 3−𝑥
(14)lim
𝑥4 − 1
(10)lim
𝑥→1 𝑥 − 1
𝑥−2
𝑥→2 √𝑥 2
−4
(B)
√1 + 𝑥 + 𝑥 2 − 1
√1 + 𝑥 − 1
√4 + 𝑥 − 2
(1)lim
(2)lim
(3)lim
𝑥→0
𝑥→0
𝑥→0
𝑥
𝑥
𝑥
√2𝑥 + 1 − 3
3
√𝑥 − 1
(4)lim
(5)lim
𝑥→4 √𝑥 − 2 − √2
𝑥→1 √𝑥 − 1
𝑚
𝑚
√𝑥 − √𝑎
(6)lim
𝑥→1 √𝑥 − 𝑎
1
1
√𝑥 2 + 2
(7) lim
(8)lim
− [𝐻𝑖𝑛𝑡: 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑐𝑜𝑛𝑗𝑢𝑔𝑎𝑡𝑒]
𝑥→∞ 3𝑥 − 6
𝑥→1 𝑥 + 1
2
29
3𝑥 − 3−𝑥
(9) 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡: lim 𝑥
= 1 [𝐻𝑖𝑛𝑡: 𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 3𝑥 ]
−𝑥
𝑥→∞ 3 − 3
(10) lim
𝑥−1
𝑥→1 √𝑥 2
(11) lim (𝑥 2 − 4𝑥)
𝑥→∞
+3−2
(C)
12 𝑥
6 −3𝑥+8
1
(1) lim [1 − ] (2) lim [1 − ]
(3) lim [1 + ]
𝑥→∞
𝑥→∞
𝑥→∞
𝑥
𝑥
𝑥
(4)lim[1 − 3𝑥 ]
𝑥→0
7
𝑥
(5) lim [
𝑥−5 2𝑥
𝑥→∞ 𝑥−1
]
(6)lim[1 − 2𝑥 ]
2𝑥−5
3
9
𝑥
𝑥→0
Right and left limits)‫(النهايات اليمنى واليسرى‬:
If the value of 𝑓(𝑥 ) approaches to the number 𝐿1 ,as x approaches
to 𝑥0 where 𝑥 > 𝑥0 from the right hand side, we write
lim 𝑓(𝑥 ) = 𝐿1
𝑥→𝑥0+
And if the value of 𝑓(𝑥 ) approaches to the number 𝐿2 as x
approaches to 𝑥0 where 𝑥 < 𝑥0 from the left hand side, we write
lim 𝑓(𝑥 ) = 𝐿2
𝑥→𝑥0−
Then if lim+𝑓(𝑥 ) = lim−𝑓(𝑥 ) = 𝐿 the limit exist and equal L. And
𝑥→𝑥0
𝑥→𝑥0
if not equal then the limit not exist)‫(غير موجوده‬.
Example:
Research)‫ (ابحث‬the limit of
30
3−𝑥 ∶𝑥 <1
(1 ) 𝑓 (𝑥 ) = { 2
𝑥 +1 ∶𝑥 >1
1−𝑥 ∶𝑥 >2
(2 ) 𝑓 (𝑥 ) = {
2𝑥 + 5 ∶ 𝑥 < 2
Solution:
3−𝑥 ∶𝑥 <1
(1 ) 𝑓 (𝑥 ) = { 2
𝑥 +1 ∶𝑥 >1
⟹ lim+𝑓(𝑥 ) = lim 3 − 𝑥 = 2 𝑎𝑛𝑑 lim−𝑓(𝑥 ) = lim 𝑥 2 + 1 = 2
𝑥→𝑥0
𝑥→1
𝑥→𝑥0
𝑥→1
⟹ lim+𝑓(𝑥 ) = lim−𝑓(𝑥 )
𝑥→𝑥0
𝑥→𝑥0
∴ 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡 𝑒𝑥𝑖𝑠𝑡 𝑎𝑛𝑑
lim 𝑓(𝑥 ) = 2
𝑥→𝑥0
1−𝑥 ∶𝑥 >2
(2 ) 𝑓 (𝑥 ) = {
2𝑥 + 5 ∶ 𝑥 < 2
⟹ lim+𝑓(𝑥 ) = lim 1 − 𝑥 = −1 𝑎𝑛𝑑 lim−𝑓(𝑥 ) = lim 2𝑥 + 5 = 9
𝑥→𝑥0
𝑥→2
𝑥→𝑥0
𝑥→2
⟹ lim+𝑓(𝑥 ) ≠ lim−𝑓(𝑥 )
𝑥→𝑥0
𝑥→𝑥0
The limit not exist
Continuity)‫(االستمراريه‬:
A function 𝑓(𝑥 ) is said to be continuous at point c if satisfied)‫(تحقق‬:
(1) 𝑓(𝑐 ) 𝑖𝑠 𝑑𝑒𝑓𝑖𝑛𝑒𝑑
(2) lim𝑓(𝑥 ) 𝑖𝑠 𝑒𝑥𝑖𝑠𝑡
𝑥→𝑐
(2) lim𝑓(𝑥 ) = 𝑓(𝑐 )
𝑥→𝑐
31
Example:
Research the continuity of the functions:
3𝑥 − 1 ∶ 𝑥 > 1
(1) 𝑓(𝑥 ) = {2
∶ 𝑥=1
𝑥2 + 1 ∶ 𝑥 < 1
𝑥 2 −𝑥−6
(3 ) 𝑓 (𝑥 ) = {
𝑥−3
4
2𝑥 2 − 1 ∶ 𝑥 > 3
(2) 𝑓(𝑥 ) = {17
∶ 𝑥=3
3𝑥
∶𝑥<3
𝑠𝑖𝑛𝑥
∶𝑥≠3
∶𝑥=3
(4 ) 𝑓 (𝑥 ) = {
𝑥
1
∶𝑥≠0
∶𝑥=0
Solution:
3𝑥 − 1 ∶ 𝑥 > 1
(1) 𝑓(𝑥 ) = {2
∶ 𝑥=1
𝑥2 + 1 ∶ 𝑥 < 1
⟹ 𝑓(1) = 2 𝑒𝑥𝑖𝑠𝑡
⟹ lim+𝑓(𝑥 ) = lim 3𝑥 − 1 = 2 𝑎𝑛𝑑 lim−𝑓(𝑥 ) = lim 𝑥 2 + 1 = 2
𝑥→𝑥0
𝑥→1
𝑥→𝑥0
𝑥→1
⟹ lim+𝑓(𝑥 ) = lim−𝑓(𝑥 ) = 2
𝑥→𝑥0
𝑥→𝑥0
⟹ lim 𝑓(𝑥 ) = 𝑓(𝑥 ) = 2
𝑥→𝑥0
The function 𝑓(𝑥 ) is continuous
----------------------------------------------------------------------------------------2𝑥 2 − 1 ∶ 𝑥 > 3
(2) 𝑓(𝑥 ) = {17
∶ 𝑥=3
3𝑥
∶𝑥<3
⟹ 𝑓(3) = 17 𝑒𝑥𝑖𝑠𝑡
32
⟹ lim+𝑓(𝑥 ) = lim 2𝑥 2 − 1 = 17 𝑎𝑛𝑑 lim−𝑓(𝑥 ) = lim 3𝑥 = 9
𝑥→𝑥0
𝑥→3
𝑥→𝑥0
𝑥→3
⟹ lim+𝑓(𝑥 ) ≠ lim−𝑓(𝑥 )
𝑥→𝑥0
𝑥→𝑥0
The limit not exist so 𝑓(𝑥 ) is not continuous
----------------------------------------------------------------------------------------𝑥2 − 𝑥 − 6
∶𝑥≠3
(3 ) 𝑓 (𝑥 ) = { 𝑥 − 3
4
∶𝑥=3
⟹ 𝑓(3) = 4 𝑒𝑥𝑖𝑠𝑡
(𝑥 − 2)(𝑥 − 3)
𝑥2 − 𝑥 − 6
⟹ lim 𝑓(𝑥 ) = lim
= lim
=1
𝑥→3
𝑥→3
(𝑥 − 3 )
𝑥−3
𝑥→𝑥0
The limit 𝑓(𝑥 ) is exist
But 𝑓(𝑥 ) is not equal to the limit 𝑓(𝑥 ) so 𝑓(𝑥 ) is not continuous
----------------------------------------------------------------------------------------𝑠𝑖𝑛𝑥
(3 ) 𝑓 (𝑥 ) = { 𝑥
1
∶𝑥≠0
∶𝑥=0
⟹ 𝑓(0) = 1 𝑒𝑥𝑖𝑠𝑡
𝑠𝑖𝑛𝑥
=1
𝑥→0 𝑥
⟹ lim 𝑓(𝑥 ) = lim
𝑥→𝑥0
The limit 𝑓(𝑥 ) is exist
lim 𝑓(𝑥 ) = 𝑓(𝑥 ) = 1 ⟹
𝑥→𝑥0
33
The function 𝑓(𝑥 ) is continuous
Example:
Find k which make)‫ ) 𝑥(𝑓 (تجعل‬continuous
𝑥 3 − 3𝑥 2
𝑓 (𝑥 ) = { 𝑥 − 3
𝑘
∶𝑥≠3
∶𝑥=3
Solution:
𝑥 3 − 3𝑥 2
𝑓 (𝑥 ) = { 𝑥 − 3
𝑘
∶𝑥≠3
∶𝑥=3
𝑥 3 − 3𝑥 2
𝑥 2 (𝑥 − 3)
⟹ lim 𝑓(𝑥 ) = lim
= lim
=9
𝑥→3
𝑥→3 𝑥 − 3
𝑥→3 (𝑥 − 3)
𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 lim 𝑓(𝑥 ) = 𝑓(3) = 9 ⟹ 𝑘 = 9
𝑥→3
Exercise:
(A)
Research the continuity of the functions:
𝑥+1 ∶𝑥 ≥1
(1 ) 𝑓 (𝑥 ) = { 2
𝑥
∶𝑥<1
2
𝑡
∶𝑡≥0
(3 ) 𝑓 (𝑥 ) = {
𝑡−2 ∶𝑡 <0
7𝑥 − 2 ∶ 𝑥 ≤ 1
(2 ) 𝑓 (𝑥 ) = { 2
2𝑥 + 3 ∶ 𝑥 > 1
𝑥2 − 9
∶𝑥>3
𝑥
+
3
(4 ) 𝑓 (𝑥 ) = {
0
∶ 𝑥=3
𝑥−3 ∶𝑥 <3
𝑥2 − 5 ∶ 𝑥 ≤ 3
𝑥 2 − 2𝑥 + 1 ∶ 𝑥 < 2
(5) 𝑓(𝑥 ) = {4
∶ 𝑥 = 3 (6 ) 𝑓 (𝑥 ) = { 2
𝑥 + 3𝑥 − 1 ∶ 𝑥 > 2
√𝑥 + 13 ∶ 𝑥 > 3
34
‫‪Find k which is make 𝑓(𝑥 ) continuous:‬‬
‫‪7𝑥 − 2 ∶ 𝑥 ≤ 1‬‬
‫‪(2 ) 𝑓 (𝑥 ) = { 2‬‬
‫𝑥𝑘‬
‫‪∶𝑥>1‬‬
‫)‪(B‬‬
‫‪2‬‬
‫𝑥𝑘‬
‫‪∶𝑥≤2‬‬
‫{ = ) 𝑥( 𝑓 ) ‪(1‬‬
‫‪2𝑥 + 𝑘 ∶ 𝑥 > 2‬‬
‫‪𝑥 2 − 16‬‬
‫‪(3 ) 𝑓 (𝑥 ) = { 𝑥 − 4 ∶ 𝑥 ≠ 4‬‬
‫𝑘‬
‫‪∶𝑥=4‬‬
‫𝑥‪𝑠𝑖𝑛5‬‬
‫‪(4) 𝑓(𝑥 ) = { 2𝑥 ∶ 𝑥 ≠ 0‬‬
‫𝑘‬
‫‪∶𝑥=0‬‬
‫ولرب نازلة يضيق بها الفتى‬
‫ذرعا وعند هللا منها المخرج‬
‫ضاقت فلما استحكمت حلقاتها‬
‫فرجت وكنت اظنها ال تفرج‬
‫‪The darkest hour is that before the dawn‬‬
‫‪35‬‬
Differentiation)‫(التفاضل‬
The derivative)‫(االشتقاق‬:
The derivative of function 𝑦 = 𝑓(𝑥 ) with respect to)‫ (بالنسبه الى‬x is
the rate)‫ (معدل‬of change)‫ (التغير‬of y with respect to x written as
𝑑𝑦
, 𝑓̀(𝑥 ), 𝑦, …
̀ and defined by the formula)‫(القاعده‬:
𝑑𝑥
𝑑𝑦
𝑓(𝑥 + ∆𝑥 ) − 𝑓(𝑥)
= lim
𝑑𝑥 ∆𝑥→0
∆𝑥
Example:
Find
𝑑𝑦
𝑑𝑥
by using definition)‫(التعريف‬
(1 ) 𝑓 (𝑥 ) = 𝑥 + 1
(2 ) 𝑓 (𝑥 ) = 𝑥 2 − 3
(3)𝑓(𝑥 ) = √𝑥
(4)𝑓(𝑥 ) = 𝑠𝑖𝑛𝑥
Solution:
(1 ) 𝑦 = 𝑥 + 1
𝑦 + ∆𝑦 = (𝑥 + ∆𝑥 ) + 1
⟹ ∆𝑦 = (𝑥 + ∆𝑥 ) + 1 − (𝑥 + 1) = ∆𝑥
⟹
∆𝑦 Δ𝑥
Δ𝑦
Δ𝑥
=
⟹ lim
= lim
= lim 1 = 1
∆𝑥→0 Δ𝑥
∆𝑥→0 Δ𝑥
∆𝑥→0
∆𝑥 Δ𝑥
𝑑𝑦
=1
𝑑𝑥
----------------------------------------------------------------------------------------(2 ) 𝑓 (𝑥 ) = 𝑥 2 − 3
36
𝑦 = 𝑥2 − 3
− − − − − − − (1)
𝑦 + ∆𝑦 = (𝑥 + ∆𝑥 )2 − 3 − − − (2)
(2) − (1) ⟹ ∆𝑦 = (𝑥 + ∆𝑥)2 − 3 − (𝑥 2 − 3 )
⟹ ∆𝑦 = 2𝑥∆𝑥 + ∆𝑥 2
∆𝑦 2𝑥∆𝑥 + ∆𝑥 2
Δ𝑦
2𝑥∆𝑥 + ∆𝑥 2
⟹
=
⟹ lim
= lim
∆𝑥→0 Δ𝑥
∆𝑥→0
∆𝑥
Δ𝑥
Δ𝑥
(2𝑥 + ∆𝑥 )
∆𝑥(2𝑥 + ∆𝑥)
= lim
= 2𝑥
∆𝑥→0
∆𝑥→0
Δ𝑥
1
= lim
-------------------------------------------------------------------------------------(3)𝑓(𝑥 ) = √𝑥
𝑦 = √𝑥 − − − − − −(1)
𝑦 + ∆𝑦 = √𝑥 + ∆𝑥 − − − −(2)
(2) − (1) ⟹ ∆𝑦 = √𝑥 + ∆𝑥 − √𝑥
𝑑𝑦
Δ𝑦
√𝑥 + ∆𝑥 − √𝑥
= lim
= lim
𝑑𝑥 ∆𝑥→0 Δ𝑥 ∆𝑥→0
Δ𝑥
(√𝑥 + ∆𝑥 − √𝑥)(√𝑥 + ∆𝑥 + √𝑥)
𝑑𝑦
= lim
𝑑𝑥 ∆𝑥→0
(Δ𝑥 )(√𝑥 + ∆𝑥 + √𝑥)
= lim
𝑥 + ∆𝑥 − 𝑥
∆𝑥→0 (Δ𝑥 )(√𝑥
+ ∆𝑥 + √𝑥)
= lim
∆𝑥→0 (√𝑥
𝑑𝑦
1
=
𝑑𝑥 2√𝑥
37
1
++ √𝑥)
=
1
2√𝑥
----------------------------------------------------------------------------------------(4)𝑓(𝑥 ) = 𝑠𝑖𝑛𝑥
𝑦 = 𝑠𝑖𝑛𝑥 − − − − − − − (1)
𝑦 + ∆𝑦 = 𝑠𝑖𝑛(𝑥 + ∆𝑥 ) − − − − − − − (2)
(2) − (1) ⟹ ∆𝑦 = 𝑠𝑖𝑛(𝑥 + ∆𝑥) − 𝑠𝑖𝑛𝑥
𝑑𝑦
Δ𝑦
𝑠𝑖𝑛(𝑥 + ∆𝑥 ) − 𝑠𝑖𝑛𝑥
= lim
= lim
𝑑𝑥 ∆𝑥→0 Δ𝑥 ∆𝑥→0
Δ𝑥
𝑠𝑖𝑛𝑥𝑐𝑜𝑠∆𝑥 + 𝑐𝑜𝑠𝑥𝑠𝑖𝑛∆𝑥 − 𝑠𝑖𝑛𝑥
∆𝑥→0
∆𝑥
= lim
𝑠𝑖𝑛𝑥(𝑐𝑜𝑠∆𝑥 − 1) + 𝑐𝑜𝑠𝑥(𝑠𝑖𝑛∆𝑥)
= lim
∆𝑥→0
∆𝑥
= 𝑠𝑖𝑛𝑥 lim
∆𝑥→0
−(1 − 𝑐𝑜𝑠∆𝑥 )
𝑠𝑖𝑛∆𝑥
+ 𝑐𝑜𝑠𝑥 lim
= 0 + 𝑐𝑜𝑠𝑥
∆𝑥→0 ∆𝑥
∆𝑥
𝑑𝑦
= 𝑐𝑜𝑠𝑥
𝑑𝑥
Exercise:
𝑑𝑦
Find 𝑑𝑥 from definition if:
(1) 𝑦 = 4𝑥 + 1
(2) 𝑦 = 𝑥 2 + 2𝑥 − 3 (3) 𝑦 = √𝑥
(4) 𝑦 = √1 + 2𝑥
(5 ) 𝑦 =
(7 ) 𝑦 =
(1 + 2𝑥 )
(1 − 2𝑥 )
1
√𝑥
(6 )
𝑦=
1
𝑥2
(8) 𝑦 = 𝑠𝑖𝑛ℎ𝑥 (9) 𝑦 = 𝑐𝑜𝑠ℎ𝑥
38
Derivative laws)‫(قوانين االشتقاق‬:
(1)Derivative of constant functions)‫(الداله الثابتة‬:
If 𝑓(𝑥 ) has the constant value 𝑓(𝑥 ) = 𝑐 then:
𝑑𝑓 𝑑𝑐
=
=0
𝑑𝑥 𝑑𝑥
Example:
Find the derivative of:
(1 )𝑦 = 3
(2) 𝑦 = −45
Solution:
(1 )𝑦 = 3
⟹
(2) 𝑦 = −45
𝑑𝑦
=0
𝑑𝑥
⟹
𝑑𝑦
=0
𝑑𝑥
(2) Derivative of Power)‫ (القوه‬rule:
If n is positive integer then:
𝑑𝑥 𝑛
𝑑[𝑓(𝑥 )]𝑛
𝑛−1
̀
(𝑖 )
(𝑖𝑖 )
= 𝑛𝑥
= 𝑛[𝑓(𝑥 )]𝑛−1 𝑓(𝑥)
𝑑𝑥
𝑑𝑥
(3)Derivative of Constant multiple rule:
If 𝑓(𝑥 ) is differentiable function of x, and c is a constant, then:
𝑑[𝑐𝑓(𝑥 )]
𝑑𝑓(𝑥)
=𝑐
𝑑𝑥
𝑑𝑥
39
(4)Derivative of Sum)‫ (الجمع‬rule:
If 𝑓(𝑥 ) 𝑎𝑛𝑑 𝑔(𝑥) are differentiable functions of x then:
𝑑 [𝑓(𝑥 ) ± 𝑔(𝑥)] 𝑑𝑓(𝑥) 𝑑𝑔(𝑥)
=
±
𝑑𝑥
𝑑𝑥
𝑑𝑥
Example:
Find the derivative of:
(1) y = x 3 + 4x − 2
(2) y = 2x −5 − 7x 4 + x −1 − 3
Solution:
(1) y = x 3 + 4x − 2 ⟹
𝑑𝑦
= 3𝑥 2 + 4
𝑑𝑥
(2) y = 2x −5 − 7x 4 + x −1 − 3 ⟹
𝑑𝑦
= −10𝑥 −6 − 28𝑥 3 − 𝑥 −2
𝑑𝑥
(5)Derivative of Product)‫ (الضرب‬rule:
If 𝑓(𝑥 ) 𝑎𝑛𝑑 𝑔(𝑥) are differentiable functions of x then:
𝑑 [𝑓(𝑥 ) ∙ 𝑔(𝑥)]
𝑑𝑔(𝑥 )
𝑑𝑓(𝑥)
= 𝑓 (𝑥 ) ∙
+ 𝑔(𝑥) ∙
𝑑𝑥
𝑑𝑥
𝑑𝑥
Example:
2
1
If u = x + 3x v = x find (1)
d(u∙v)
dx
(2 )
d(u+v)
dx
u
(3 )
d(v)
(6)Derivative of Quotient )‫(القسمه‬rule:
If 𝑓(𝑥 ) 𝑎𝑛𝑑 𝑔(𝑥) are differentiable functions of x then:
40
dx
𝑑[
𝑓(𝑥)
] 𝑔(𝑥 ) ∙ 𝑓(̀𝑥 ) − 𝑓(𝑥) ∙ 𝑔̀ (𝑥)
𝑔(𝑥)
=
(𝑔(𝑥))2
𝑑𝑥
Example:
1
If u = x 2 + 3x , v = x find (1)
d(u∙v)
dx
(2 )
d(u+v)
dx
u
(3 )
d(v)
dx
Solution:
(1 )
d (u ∙ v )
𝑑𝑣
𝑑𝑢
−1
1
=𝑢∙
+𝑣∙
= (x 2 + 3x) ( 2 ) + ( ) (2𝑥 + 3)
dx
𝑑𝑥
𝑑𝑥
𝑥
x
(2 )
d(u + v) 𝑑𝑢 𝑑𝑣
1
=
+
= 2𝑥 + 3 − 2
dx
𝑑𝑥 𝑑𝑥
𝑥
(3 )
u
d (v)
dx
=
v∙
1
−1
𝑑𝑢
𝑑𝑣
( x ) (2𝑥 + 3) − (x 2 + 3x) ( 2 )
−𝑢∙
𝑑𝑥
𝑑𝑥 =
𝑥
2
v2
1
(x)
(7)The chain )‫(السلسله‬rule:
If the function 𝑦 = 𝑓(𝑢) is differentiable at 𝑢 = 𝑔(𝑥) and 𝑔(𝑥) is
differentiable functions of x then:
𝑑𝑦 𝑑𝑓 𝑑𝑓 𝑑𝑢
=
=
∙
𝑑𝑥 𝑑𝑥 𝑑𝑢 𝑑𝑥
Example:
If 𝑦 = 𝑢2 + 5𝑢 𝑎𝑛𝑑 𝑢 = 2𝑥 3 − 𝑥 𝑓𝑖𝑛𝑑
41
𝑑𝑦
𝑑𝑥
Solution:
𝑑𝑦 𝑑𝑦 𝑑𝑢
=
∙
𝑑𝑥 𝑑𝑢 𝑑𝑥
𝑑𝑦
= 2𝑢 + 5 𝑎𝑛𝑑
𝑑𝑢
𝑑𝑢
= 6𝑥 2 − 1
𝑑𝑥
𝑑𝑦
⟹
= (2𝑢 + 5 ) ∙ (6𝑥 2 − 1)
𝑑𝑥
Exercise:
𝑑𝑦
Find 𝑑𝑥 if
(1) y = x 4 + 3x + 2
x 3 + 3x + 1
(3 ) y =
x4 + 3
5
(5) 𝑦 = 𝑥 + 4𝑥
−3
(2) y = (4x 2−1 )(7x 3 + x)
(4 ) 𝑦 =
1
√𝑥 + 2𝑥
2
− 10𝑥 + 7 (6) 𝑦 =
(7) 𝑦 = √1 + √𝑥 (8) 𝑦 = 2𝑥 2 √2 − 𝑥
(10) 𝑦 =
𝑥
√2 + 3𝑥 2
1
𝑥2
+ 7𝑥 −2
(9 ) 𝑦 = √
𝑥−1
𝑥+1
7
𝑥3 + 1
(11) 𝑦 = ( 2
)
5𝑥 − 1
(12) 𝑦 =
2
5
+3
√𝑥 √𝑥
(8)Derivative of Parametric)‫ (البارمتريه‬equations:
A parameterized equation 𝑥 = 𝑓 (𝑡) 𝑎𝑛𝑑 𝑦 = 𝑔(𝑡)
Is differentiable at t if 𝑓(𝑡) 𝑎𝑛𝑑 𝑔(𝑡) are differentiable at t then:
42
𝑑𝑦 𝑑𝑦 𝑑𝑥
=
÷
𝑑𝑥 𝑑𝑡 𝑑𝑡
For all derivative exist and
𝑑𝑥
𝑑𝑡
≠0
(9)Derivative of Exponential functions:
If 𝑓(𝑥 ) = 𝑎𝑢(𝑥) then:
𝑑𝑦 𝑑 [𝑎𝑢 ]
𝑑𝑢
=
= (𝑎𝑢(𝑥) ) ( ) (lna)
𝑑𝑥
𝑑𝑥
𝑑𝑥
Note:
If 𝑎 = 𝑒 then 𝑓(𝑥 ) = 𝑒 𝑢(𝑥) ⟹
𝑑𝑦
𝑑𝑥
=
𝑑[𝑒 𝑢 ]
𝑑𝑥
𝑑𝑢
= (𝑒 𝑢(𝑥) ) ( )
𝑑𝑥
(10)Derivative of Logarithmic functions:
If 𝑓(𝑥 ) = log 𝑎 𝑢(𝑥) then:
𝑑𝑦 𝑑 [log 𝑎 𝑢(𝑥)]
1
𝑑𝑢
=
=(
) ( ) (log 𝑎 𝑒)
𝑑𝑥
𝑑𝑥
𝑢(𝑥) 𝑑𝑥
Note:
If 𝑎 = 𝑒 then 𝑓(𝑥 ) = log 𝑒 𝑢(𝑥 ) = 𝑙𝑛𝑢(𝑥 )
𝑑𝑦
1
𝑑𝑢
⟹
=(
)( )
𝑑𝑥
𝑢(𝑥) 𝑑𝑥
Example:
(1) 𝑦 = 35𝑥
(2 ) 𝑦 = 7 𝑥
2 +3
(3) 𝑦 = 𝑒 𝑠𝑖𝑛𝑥
(4) 𝑦 = 𝑙𝑛(𝑥 2 + 7) (5) 𝑦 = 𝑙𝑛(𝑐𝑜𝑠8𝑥) (6)𝑦 = 𝑙𝑛(513𝑥 )
43
Solution:
(1) 𝑦 = 35𝑥
(2 ) 𝑦 = 7
𝑥 2 +3
(3) 𝑦 = 𝑒 𝑠𝑖𝑛𝑥
⟹
𝑑𝑦
= (35𝑥 )(5)(𝑙𝑛3)
𝑑𝑥
𝑑𝑦
2
⟹
= (7𝑥 +3 )(2𝑥)(𝑙𝑛7)
𝑑𝑥
𝑑𝑦
⟹
= (𝑐𝑜𝑠𝑥)𝑒 𝑠𝑖𝑛𝑥
𝑑𝑥
(4) 𝑦 = 𝑙𝑛(𝑥 2 + 7)
𝑑𝑦
1
⟹
=( 2
) (2𝑥 )
𝑑𝑥
𝑥 +7
(5) 𝑦 = 𝑙𝑛(𝑐𝑜𝑠8𝑥 ) ⟹
(6)𝑦 = 𝑙𝑛(513𝑥 ) ⟹
𝑑𝑦
1
=(
) (−8𝑠𝑖𝑛8𝑥 )
𝑑𝑥
𝑐𝑜𝑠8𝑥
𝑑𝑦
1
= ( 13𝑥 ) [(513𝑥 )(13)(𝑙𝑛5)] = 13𝑙𝑛5
𝑑𝑥
5
Example:
𝑑𝑦
Find 𝑑𝑥 if:
(1)𝑦 = log 3 (𝑥 2 + 1) + 𝑥 2
(2)𝑦 = 𝑙𝑛(𝑙𝑛𝑥) + 𝑒 −2𝑥
(3)𝑦 = 𝑥 3 𝑙𝑛(5 − 2𝑥) (4)𝑦 = 𝑙𝑛 (
𝑥
√2𝑥+1 + 2 𝑥
(
)
)
5
𝑦
=
𝑒
1 + 𝑥2
(6)𝑦 = 𝑙𝑛(𝑥 + 3)(2𝑥 − 1)
Solution:
(1)𝑦 = log 3 (𝑥 2 + 1) + 𝑥 2 ⟹
𝑑𝑦
1
=( 2
) (2𝑥)(log 3 𝑒) + 2𝑥
𝑑𝑥
𝑥 +1
44
(2)𝑦 = 𝑙𝑛(𝑙𝑛𝑥) + 𝑒 −2𝑥 ⟹
(3)𝑦 = 𝑥 3 𝑙𝑛(5 − 2𝑥) ⟹
𝑑𝑦
1
1
= ( ) [( ) (1)] − 2𝑒 −2𝑥
𝑑𝑥
𝑙𝑛𝑥
𝑥
𝑑𝑦
−2
= 𝑥3 [
] + (3𝑥 2 )𝑙𝑛(5 − 2𝑥 )
𝑑𝑥
5 − 2𝑥
𝑥
𝑑𝑦
1 + 𝑥 2 (1 + 𝑥 2 )(1) − (𝑥 )(2𝑥)
(4)𝑦 = 𝑙𝑛 (
) ⟹
=[
][
]
(1 + 𝑥 2 )2
1 + 𝑥2
𝑑𝑥
𝑥
(5)𝑦 = 𝑒 √2𝑥+1 + 2𝑥
−1
𝑑𝑦
1
⟹
= [ (2𝑥 + 1) 2 (2)] 𝑒 √2𝑥+1 + (2𝑥 )(1)(𝑙𝑛2)
𝑑𝑥
2
(6)𝑦 = 𝑙𝑛(𝑥 + 3)(2𝑥 − 1) = 𝑙𝑛(2𝑥 2 − 5𝑥 − 3)
𝑑𝑦
4𝑥 − 5
⟹
=
𝑑𝑥 (2𝑥 2 − 5𝑥 − 3)
Example:
(1)𝑦 = (𝑥 2 + 1)(𝑥 3 − 2)(4𝑥 − 1)
(2)𝑦 = (𝑥 + 5)7 (2 + 6𝑥)3
Solution:
(1)𝑦 = (𝑥 2 + 1)(𝑥 3 − 2)(4𝑥 − 1)
⟹ 𝑙𝑛𝑦 = 𝑙𝑛(𝑥 2 + 1)(𝑥 3 − 2)(4𝑥 − 1)
= 𝑙𝑛(𝑥 2 + 1) + 𝑙𝑛(𝑥 3 − 2) + 𝑙𝑛(4𝑥 − 1)
1 𝑑𝑦
2𝑥
3𝑥 2
4
⟹( )
=( 2
)+( 3
)+(
)
𝑦 𝑑𝑥
𝑥 +1
𝑥 −2
4𝑥 − 1
45
𝑑𝑦
2𝑥
3𝑥 2
4
⟹
= [( 2
)+( 3
)+(
)] 𝑦
𝑑𝑥
𝑥 +1
𝑥 −2
4𝑥 − 1
𝑑𝑦
2𝑥
3𝑥 2
4
⟹
= [( 2
)+( 3
)+(
)] ((𝑥 2 + 1)(𝑥 3 − 2)(4𝑥 − 1))
𝑑𝑥
𝑥 +1
𝑥 −2
4𝑥 − 1
-----------------------------------------------------------------------------------------------------
(2)𝑦 = (𝑥 + 5)7 (2 + 6𝑥)3 ⟹ 𝑙𝑛𝑦 = 𝑙𝑛(𝑥 + 5)7 (2 + 6𝑥 )3
⟹ 𝑙𝑛𝑦 = 7 ln(𝑥 + 5) + 3ln(2 + 6𝑥)
1 𝑑𝑦
7
18
⟹( )
=(
)+(
)
𝑦 𝑑𝑥
𝑥+5
2 + 6𝑥
⟹
𝑑𝑦
7
18
= [(
)+(
)] [(𝑥 + 5)7 (2 + 6𝑥 )3 ]
𝑑𝑥
𝑥+5
2 + 6𝑥
Example:
𝑑𝑦
Find 𝑑𝑥 if:
(1 ) 𝑦 = 𝑥
𝑥
(4 ) 𝑦 = 𝑥 𝑒
𝑥
2
(2 ) 𝑦 = (𝑥 + 3
)𝑙𝑛𝑥
(3) 𝑦 = (5 −
1
𝑥
𝑥)
(5) 𝑦 = (𝑠𝑖𝑛𝑥 )𝑥 (6) 𝑦 = 𝑥 𝑡𝑎𝑛𝑥
Solution
(1 ) 𝑦 = 𝑥 𝑥
⟹ 𝑙𝑛𝑦 = ln 𝑥 𝑥 = 𝑥𝑙𝑛𝑥
1 𝑑𝑦
1
𝑑𝑦
⟹( )
= 𝑥 ( ) + 𝑙𝑛𝑥 ⟹
= (1 + 𝑙𝑛𝑥)𝑦 = (1 + 𝑙𝑛𝑥)𝑥 𝑥
𝑦 𝑑𝑥
𝑥
𝑑𝑥
-----------------------------------------------------------------------------------------
46
(2) 𝑦 = (𝑥 2 + 3)𝑙𝑛𝑥 ⟹ 𝑙𝑛𝑦 = 𝑙𝑛(𝑥 2 + 3)𝑙𝑛𝑥 = 𝑙𝑛𝑥(𝑙𝑛(𝑥 2 + 3))
1 𝑑𝑦
2𝑥
1
⟹( )
= 𝑙𝑛𝑥 [ 2
] + ( ) (𝑙𝑛(𝑥 2 + 3))
(𝑥 + 3 )
𝑦 𝑑𝑥
𝑥
⟹
𝑑𝑦
2𝑥
1
= [𝑙𝑛𝑥 [ 2
] + ( ) (𝑙𝑛(𝑥 2 + 3))] (𝑥 2 + 3)𝑙𝑛𝑥
(𝑥 + 3 )
𝑑𝑥
𝑥
----------------------------------------------------------------------------------------(3 ) 𝑦 = (5 −
1
𝑥 )𝑥
⟹ 𝑙𝑛𝑦 = 𝑙𝑛(5 −
1
𝑥 )𝑥
1
= ( ) ln(5 − 𝑥)
𝑥
1 𝑑𝑦
1
−1
−1
⟹( )
= ( )[
] + (5 − 𝑥 ) ( 2 )
𝑦 𝑑𝑥
𝑥 5−𝑥
𝑥
1
𝑑𝑦
1
−1
−1
𝑥
⟹
= [( ) [
] + (5 − 𝑥 ) ( 2 )] (5 − 𝑥 )
𝑑𝑥
𝑥 5−𝑥
𝑥
----------------------------------------------------------------------------------------(4 ) 𝑦 = 𝑥 𝑒
𝑥
𝑥
⟹ 𝑙𝑛𝑦 = 𝑙𝑛𝑥 𝑒 = 𝑒 𝑥 (𝑙𝑛𝑥)
1 𝑑𝑦
1
⟹( )
= 𝑒 𝑥 ( ) + 𝑒 𝑥 (𝑙𝑛𝑥)
𝑦 𝑑𝑥
𝑥
⟹
𝑑𝑦
1
𝑥
= [𝑒 𝑥 ( ) + 𝑒 𝑥 (𝑙𝑛𝑥)] 𝑥 𝑒
𝑑𝑥
𝑥
----------------------------------------------------------------------------------------(5) 𝑦 = (𝑠𝑖𝑛𝑥 )𝑥
⟹ 𝑙𝑛𝑦 = 𝑙𝑛(𝑠𝑖𝑛𝑥 )𝑥 = 𝑥(𝑙𝑛𝑠𝑖𝑛𝑥 )
1 𝑑𝑦
𝑐𝑜𝑠𝑥
⟹( )
= 𝑥(
) + 𝑠𝑖𝑛𝑥
𝑦 𝑑𝑥
𝑠𝑖𝑛𝑥
47
⟹
𝑑𝑦
𝑐𝑜𝑠𝑥
= [𝑥 (
) + 𝑠𝑖𝑛𝑥] (𝑠𝑖𝑛𝑥 )𝑥
𝑑𝑥
𝑠𝑖𝑛𝑥
----------------------------------------------------------------------------------------(6) 𝑦 = 𝑥 𝑡𝑎𝑛𝑥 ⟹ 𝑙𝑛𝑦 = 𝑙𝑛𝑥 𝑡𝑎𝑛𝑥 = 𝑡𝑎𝑛𝑥 (𝑙𝑛𝑥)
1 𝑑𝑦
1
⟹( )
= (𝑡𝑎𝑛𝑥 ) ( ) + (𝑙𝑛𝑥)(𝑠𝑒𝑐 2 𝑥 )
𝑦 𝑑𝑥
𝑥
(11)Derivative of Implicit )‫(الضمني‬differentiation:
To find the Implicit differentiation we:
(i)Differentiate both sides of the equation with respect to x,
treating y as differentiable function of x.
(ii)Collect the terms with
𝑑𝑦
𝑑𝑥
on one side of the equation.
𝑑𝑦
(iii)Solve for 𝑑𝑥
Example:
(1 ) 𝑥 2 + 𝑦 2 = 6
(2) 5𝑦 2 + √𝑦 = 𝑥 2 (3) 𝑒 𝑦
2 +3𝑥
− 𝑡𝑎𝑛𝑦 = 4
(4) (𝑡𝑎𝑛𝑥 )(𝑠𝑖𝑛𝑦) = 𝑥 + 1
Solution:
(1 ) 𝑥 2 + 𝑦 2 = 6
⟹ 2𝑥 + 2𝑦
𝑑𝑦
𝑑𝑦 −𝑥
=0 ⟹
=
𝑑𝑥
𝑑𝑥
𝑦
----------------------------------------------------------------------------------------48
(2) 5𝑦 2 + √𝑦 = 𝑥 2 ⟹ 10𝑦
⟹
𝑑𝑦
=
𝑑𝑥
𝑑𝑦 1 −1 𝑑𝑦
+ 𝑦 2
= 2𝑥
𝑑𝑥 2
𝑑𝑥
2𝑥
1 1
(10𝑦 + 2 𝑦 −2 )
----------------------------------------------------------------------------------------(3 ) 𝑒 𝑦
(2𝑦𝑒 𝑦
2 +3𝑥
2 +3𝑥
− 𝑡𝑎𝑛𝑦 = 4 ⟹ (2𝑦
− 𝑠𝑒𝑐 2 𝑦)
𝑑𝑦
𝑑𝑦
2
+ 3) 𝑒 𝑦 +3𝑥 − 𝑠𝑒𝑐 2 𝑦
=0
𝑑𝑥
𝑑𝑥
𝑑𝑦
2
= −3𝑒 𝑦 +3𝑥
𝑑𝑥
2
𝑑𝑦
−3𝑒 𝑦 +3𝑥
⟹
=
𝑑𝑥 (2𝑦𝑒 𝑦2 +3𝑥 − 𝑠𝑒𝑐 2 𝑦)
----------------------------------------------------------------------------------------(4) (𝑡𝑎𝑛𝑥 )(𝑠𝑖𝑛𝑦) = 𝑥 + 1
⟹ 𝑡𝑎𝑛𝑥 (𝑐𝑜𝑠𝑦)
𝑑𝑦
+ 𝑠𝑒𝑐 2 𝑥(𝑠𝑖𝑛𝑦) = 1
𝑑𝑥
𝑑𝑦 1 − 𝑠𝑒𝑐 2 𝑥(𝑠𝑖𝑛𝑦)
⟹
=
𝑑𝑥
𝑡𝑎𝑛𝑥 (𝑐𝑜𝑠𝑦)
(12)Derivative of trigonometric function:
If 𝑢 = 𝑢(𝑥) we find
𝑑 (𝑠𝑖𝑛𝑢)
𝑑𝑢
(1 )
= 𝑐𝑜𝑠𝑢 ( )
𝑑𝑥
𝑑𝑥
49
(2 )
𝑑 (𝑐𝑜𝑠𝑢)
𝑑𝑢
= −𝑠𝑖𝑛𝑢 ( )
𝑑𝑥
𝑑𝑥
𝑑 (𝑡𝑎𝑛𝑢)
𝑑𝑢
(3 )
= 𝑠𝑒𝑐 2 𝑢 ( )
𝑑𝑥
𝑑𝑥
(4 )
𝑑 (𝑐𝑜𝑡𝑢)
𝑑𝑢
= −𝑐𝑜𝑠𝑒𝑐 2 𝑢 ( )
𝑑𝑥
𝑑𝑥
(5 )
𝑑 (𝑠𝑒𝑐𝑢)
𝑑𝑢
= (𝑠𝑒𝑐𝑢)(𝑡𝑎𝑛𝑢) ( )
𝑑𝑥
𝑑𝑥
(6 )
𝑑 (𝑐𝑜𝑠𝑒𝑐𝑢)
𝑑𝑢
(
)(
)
= − 𝑐𝑜𝑠𝑒𝑐𝑢 𝑐𝑜𝑡𝑢 ( )
𝑑𝑥
𝑑𝑥
Example:
Find
𝑑𝑦
𝑑𝑥
if
(1) 𝑦 = 𝑥 2 + 𝑠𝑖𝑛𝑥 (2) 𝑦 =
(4)y = cot5x
𝑐𝑜𝑠𝑥
(3) 𝑦 = 𝑠𝑖𝑛𝑥 (𝑡𝑎𝑛𝑥 )
𝑥
(5) y = sec8x
Solution:
(1) y = x 2 + sinx
(2 ) y =
⟹
dy
= 2x + cosx
dx
cosx
dy
= x −1 cosx ⟹
= −x −1 sinx − x −2 cosx
x
dx
(3) 𝑦 = 𝑠𝑖𝑛𝑥𝑡𝑎𝑛𝑥 ⟹
𝑑𝑦
= 𝑠𝑖𝑛𝑥(𝑠𝑒𝑐 2 𝑥 ) + 𝑐𝑜𝑠𝑥(𝑡𝑎𝑛𝑥 )
𝑑𝑥
50
(4)𝑦 = 𝑐𝑜𝑡5𝑥
⟹
(5) 𝑦 = 𝑠𝑒𝑐8𝑥 ⟹
𝑑𝑦
= −5𝑐𝑜𝑠𝑒𝑐 2 5𝑥
𝑑𝑥
𝑑𝑦
= 8(𝑠𝑒𝑐8𝑥)(𝑡𝑎𝑛8𝑥 )
𝑑𝑥
Example:
Prove the derivative of
(1)𝑡𝑎𝑛𝑥 (2) 𝑐𝑜𝑡𝑥 (3) 𝑠𝑒𝑐𝑥 (4) 𝑐𝑜𝑠𝑒𝑐𝑥
Solution:
(1)𝑦 = 𝑡𝑎𝑛𝑥 =
𝑠𝑖𝑛𝑥
𝑑𝑦 (𝑐𝑜𝑠𝑥)(𝑐𝑜𝑠𝑥) − (𝑠𝑖𝑛𝑥 )(−𝑠𝑖𝑛𝑥 )
⟹
=
(𝑐𝑜𝑠𝑥)2
𝑐𝑜𝑠𝑥
𝑑𝑥
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥
1
=
=
= 𝑠𝑒𝑐 2 𝑥
2
2
𝑐𝑜𝑠 𝑥
𝑐𝑜𝑠 𝑥
----------------------------------------------------------------------------------------𝑐𝑜𝑠𝑥
𝑑𝑦 (𝑠𝑖𝑛𝑥 )(−𝑠𝑖𝑛𝑥 ) − (𝑐𝑜𝑠𝑥)(𝑐𝑜𝑠𝑥)
(2) 𝑦 = 𝑐𝑜𝑡𝑥 =
⟹
=
(𝑠𝑖𝑛𝑥 )2
𝑠𝑖𝑛𝑥
𝑑𝑥
−(𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥)
−1
=
=
= −𝑐𝑜𝑠𝑒𝑐 2 𝑥
2
2
(𝑠𝑖𝑛𝑥 )
(𝑠𝑖𝑛𝑥)
----------------------------------------------------------------------------------------(3) 𝑦 = 𝑠𝑒𝑐𝑥 =
=
(𝑐𝑜𝑠𝑥)(0) − (1)(−𝑠𝑖𝑛𝑥 )
1
𝑑𝑦
⟹
=
(𝑐𝑜𝑠𝑥)2
𝑐𝑜𝑠𝑥
𝑑𝑥
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑥
=
= 𝑠𝑒𝑐𝑥(𝑡𝑎𝑛𝑥 )
(𝑐𝑜𝑠𝑥)2 (𝑐𝑜𝑠𝑥)(𝑐𝑜𝑠𝑥)
51
----------------------------------------------------------------------------------------(4)𝑦 = 𝑐𝑜𝑠𝑒𝑐𝑥 =
1
𝑑𝑦 (𝑠𝑖𝑛𝑥)(0) − (1)(𝑐𝑜𝑠𝑥 )
⟹
=
(𝑠𝑖𝑛𝑥 )2
𝑠𝑖𝑛𝑥
𝑑𝑥
(𝑐𝑜𝑠𝑥)
𝑐𝑜𝑠𝑥
=
=
= 𝑐𝑜𝑠𝑒𝑐𝑥(𝑐𝑜𝑡𝑥 )
(𝑠𝑖𝑛𝑥 )2 (𝑠𝑖𝑛𝑥 )(𝑠𝑖𝑛𝑥 )
(13)Derivative of Inverse trigonometric functions:
(1 )
𝑑
1
𝑑𝑢
[𝑠𝑖𝑛−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
√1 − 𝑢2 𝑑𝑥
(2 )
𝑑
−1
𝑑𝑢
[𝑐𝑜𝑠 −1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
√1 − 𝑢2 𝑑𝑥
(3 )
𝑑
1
𝑑𝑢
[𝑡𝑎𝑛−1 𝑢(𝑥)] = [
]
[
]
𝑑𝑥
1 + 𝑢2 𝑑𝑥
(4 )
𝑑
−1
𝑑𝑢
[𝑡𝑎𝑛−1 𝑢(𝑥)] = [
]
[
]
𝑑𝑥
1 + 𝑢2 𝑑𝑥
(5 )
𝑑
1
𝑑𝑢
[𝑠𝑒𝑐 −1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
(6 )
𝑑
−1
𝑑𝑢
[𝑠𝑒𝑐 −1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
Example:
Prove that:
(1 )
𝑑
1
[𝑠𝑖𝑛−1 𝑥 ] = [
]
𝑑𝑥
√1 − 𝑥 2
52
Solution:
𝑦 = 𝑠𝑖𝑛−1 𝑥 ⟹ 𝑥 = 𝑠𝑖𝑛𝑦 ⟹
𝑑𝑥
𝑑𝑦
1
= 𝑐𝑜𝑠𝑦 ⟹
=
𝑑𝑦
𝑑𝑥 𝑐𝑜𝑠𝑦
𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 = 1 ⟹ 𝑐𝑜𝑠𝑦 = √1 − 𝑠𝑖𝑛2 𝑦
⟹
𝑑𝑦
1
1
=
=
𝑑𝑥 √1 − 𝑠𝑖𝑛2 𝑦 √1 − 𝑥 2
⟹
𝑑
1
[𝑠𝑖𝑛−1 𝑥] =
𝑑𝑥
√1 − 𝑥 2
(14)Derivative of Hyperbolic functions:
If 𝑢 = 𝑢(𝑥) we find
(1 )
𝑑 (𝑠𝑖𝑛ℎ𝑢)
𝑑𝑢
= 𝑐𝑜𝑠ℎ𝑢 ( )
𝑑𝑥
𝑑𝑥
(2 )
𝑑 (𝑐𝑜𝑠ℎ𝑢)
𝑑𝑢
= 𝑠𝑖𝑛ℎ𝑢 ( )
𝑑𝑥
𝑑𝑥
(3 )
𝑑 (𝑡𝑎𝑛ℎ𝑢)
𝑑𝑢
2
= 𝑠𝑒𝑐ℎ 𝑢 ( )
𝑑𝑥
𝑑𝑥
𝑑 (𝑐𝑜𝑡ℎ𝑢)
𝑑𝑢
(4 )
= −𝑐𝑜𝑠𝑒𝑐ℎ2 𝑢 ( )
𝑑𝑥
𝑑𝑥
(5 )
𝑑 (𝑠𝑒𝑐ℎ𝑢)
𝑑𝑢
= −(𝑠𝑒𝑐ℎ𝑢)(𝑡𝑎𝑛ℎ𝑢) ( )
𝑑𝑥
𝑑𝑥
(6 )
𝑑 (𝑐𝑜𝑠𝑒𝑐𝑢)
𝑑𝑢
= −(𝑐𝑜𝑠𝑒𝑐ℎ𝑢)(𝑐𝑜𝑡ℎ𝑢) ( )
𝑑𝑥
𝑑𝑥
53
(15)Derivative of Inverse hyperbolic functions:
(1 )
𝑑
1
𝑑𝑢
[𝑠𝑖𝑛ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
√1 + 𝑢2 𝑑𝑥
(2 )
𝑑
−1
𝑑𝑢
[𝑐𝑜𝑠ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
√1 − 𝑢2 𝑑𝑥
(3 )
𝑑
1
𝑑𝑢
[𝑡𝑎𝑛ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
1 − 𝑢2 𝑑𝑥
(4 )
𝑑
−1
𝑑𝑢
[𝑡𝑎𝑛ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
1 − 𝑢2 𝑑𝑥
(5 )
𝑑
1
𝑑𝑢
[𝑠𝑒𝑐ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
(6 )
𝑑
−1
𝑑𝑢
[𝑠𝑒𝑐ℎ−1 𝑢(𝑥)] = [
][ ]
𝑑𝑥
𝑢√𝑢2 − 1 𝑑𝑥
Exercise:
(A)
(1 ) 𝑦 =
𝑢−1
𝑢+1
𝑢 = √𝑥
(3)𝑥 3 − 4𝑥𝑦 + 𝑦 2 = 12
(2 ) 𝑦 = 𝑢 3 + 4 ,
𝑢 = 𝑥 2 − 3𝑥
(4 )
‫ فاحرص عليهما‬، ‫اإلميان والعمل الصاحل مها سر حياتك الطيبة‬
54
Higher derivative)‫(المشتقات العليا‬
If 𝑦 = 𝑓(𝑥) is a derivative function then:
𝑑𝑦
𝑑𝑓
𝑦̀ = 𝑓̀(𝑥 ) = 𝑑𝑥 = 𝑑𝑥 is the first derivative of a function f
𝑑2 𝑦
𝑑2 𝑓
𝑦̀̀ = 𝑓̀ (𝑥 ) = 2 = 2 is the second derivative of a function f
𝑑𝑥
𝑑𝑥
.
.
.
𝑦 𝑛 = 𝑓 𝑛 (𝑥 ) =
𝑑𝑛 𝑦
𝑑𝑥 𝑛
=
𝑑𝑛 𝑓
𝑑𝑥 𝑛
is the nth derivative of a function f
Example:
Find the third derivative of:
(1) 𝑦 = 2𝑥 4 − 𝑥 −1 + 6𝑥
(2) 𝑦 = 𝑠𝑖𝑛2𝑥 + 𝑙𝑛𝑥
Solution:
(1) 𝑦 = 2𝑥 4 − 𝑥 −1 + 6𝑥
⟹ 𝑦̀ =
𝑑𝑦
= 8𝑥 3 + 𝑥 −2 + 6
𝑑𝑥
⟹ 𝑦̀̀ = 24𝑥 2 − 2𝑥 −3 ⟹ 𝑦̀̀ = 48𝑥 + 6𝑥 −4
----------------------------------------------------------------------------------------(2) 𝑦 = 𝑠𝑖𝑛2𝑥 + 𝑙𝑛𝑥
𝑑𝑦
1
⟹ 𝑦̀ =
= 2𝑐𝑜𝑠2𝑥 +
𝑑𝑥
𝑥
55
⟹ 𝑦̀̀ = −4𝑠𝑖𝑛2𝑥 − 𝑥 −2 ⟹ 𝑦̀̀ = −8𝑐𝑜𝑠2𝑥 + 2𝑥 −3
L’Hopital’s rule)‫(قاعدة لوبتال‬:
𝑓(𝑥)
0
If lim 𝑔(𝑥) 𝑤ℎ𝑒𝑟𝑒 𝑔(𝑥) ≠ 0 has a value 0 𝑜𝑟
𝑥→𝑎
∞
∞
then
𝑓(𝑥)
𝑓̀(𝑥)
lim
= lim
𝑥→𝑎 𝑔(𝑥)
𝑥→𝑎 𝑔̀ (𝑥)
If lim 𝑓(𝑥) ∙ 𝑔(𝑥) = 0 ∙ ∞ 𝑎𝑛𝑑 lim 𝑓(𝑥) − 𝑔(𝑥) = ∞ − ∞
𝑥→𝑎
𝑥→𝑎
We can be converted to the form
0
0
𝑜𝑟
∞
∞
Example:
Use L’Hopital’s rule to find:
(1) lim𝜋
𝑥→ 2
1 − 𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥 − 𝑥
𝑥→0
𝑥3
(4) lim
𝑙𝑛𝑥
𝑠𝑖𝑛4𝑥
(3)lim
𝑥→∞ 𝑥
𝑥→0 𝑠𝑖𝑛ℎ2𝑥
(2) lim
(5) lim𝜋(1 − 𝑡𝑎𝑛𝑥 )𝑠𝑒𝑐2𝑥
𝑥→ 4
Solution:
1 − 𝑠𝑖𝑛𝑥
0
=
0
𝑥→ 2 𝑐𝑜𝑠𝑥
(1) lim𝜋
⟹By L’Hopital’s rule
−𝑐𝑜𝑠𝑥
0
lim𝜋 −𝑠𝑖𝑛𝑥 = 1 = 0
𝑥→ 2
𝑙𝑛𝑥 ∞
=
𝑥→∞ 𝑥
∞
(2) lim
56
⟹By L’Hopital’s rule lim
1
𝑥
𝑥→∞ 1
= 𝑧𝑒𝑟𝑜
𝑠𝑖𝑛4𝑥
0
=
𝑥→0 𝑠𝑖𝑛ℎ2𝑥
0
(3)lim
4𝑐𝑜𝑠4𝑥
4
⟹By L’Hopital’s rule lim 2𝑐𝑜𝑠ℎ2𝑥 = 2 = 2
𝑥→0
𝑠𝑖𝑛𝑥 − 𝑥
0
=
𝑥→0
𝑥3
0
(4) lim
⟹By L’Hopital’s rule lim
𝑥→0
𝑐𝑜𝑠𝑥−1
3𝑥 2
⟹By L’Hopital’s rule again lim
=
0
−𝑠𝑖𝑛𝑥
6𝑥
𝑥→0
⟹By L’Hopital’s rule again lim
0
−𝑐𝑜𝑠𝑥
6
𝑥→0
0
=0
=
−1
6
(5) lim𝜋(1 − 𝑡𝑎𝑛𝑥 )𝑠𝑒𝑐2𝑥 = 0 ∙ ∞
𝑥→ 4
lim𝜋
1−𝑡𝑎𝑛𝑥
𝑥→ 4 𝑐𝑜𝑠2𝑥
⟹By L’Hopital’s rule lim𝜋
𝑠𝑒𝑐 2 𝑥
𝑥→ 4 −2𝑠𝑖𝑛2𝑥
=
−2
−2(1)
=1
Exercise:
(A)
(1) lim 𝑐𝑜𝑠𝑒𝑐𝑥 − 𝑐𝑜𝑡𝑥
𝑥→0
𝑥
𝑥→0 𝑠𝑖𝑛𝑥
𝑠𝑖𝑛4𝑥
𝑥→0 𝑠𝑖𝑛5𝑥
(2) lim
1 − 𝑐𝑜𝑠𝑥
𝑥→0
𝑥
𝑡𝑎𝑛2𝑥
𝑥→0 𝑡𝑎𝑛7
(3)lim
(4)lim
(5)lim
(6)lim 𝑥𝑙𝑛𝑥
𝑠𝑖𝑛4𝑥
(7)lim
𝑥→0 𝑡𝑎𝑛2𝑥
𝑥𝑙𝑛𝑥
(8)lim
𝑥→0 1 − 𝑥
𝑒𝑥 − 1
(9)lim
𝑥→0
𝑥
57
𝑥→0
𝑒𝑥 − 1
(10)lim
𝑥→0 𝑙𝑛(𝑥 + 1)
1 − 𝑥2
(11)lim
𝑥→1 1 − 𝑥 3
10𝑥 − 1
(12)lim
𝑥→0
𝑥
1
𝑥 + 𝑠𝑖𝑛𝜋𝑥
𝑡𝑎𝑛𝑥
(
)
(
)
− 𝑐𝑜𝑠𝑒𝑐𝑥 14 lim
15 lim
𝑥→0 𝑥
𝑥→0 𝑥 − 𝑠𝑖𝑛𝜋𝑥
𝑥→𝜋 𝑥
(13)lim
(16) lim √𝑥 2 + 𝑥 + 1 − √𝑥 2 − 𝑥
𝑥→∞
(B)
1 𝑥+3
(1) lim (1 + )
𝑥→∞
𝑥
1 5𝑥
𝑥+1 𝑥
(2) lim (1 + )
(3) lim (
)
𝑥→∞
𝑥→∞ 𝑥 − 1
𝑥
2
1
(4)lim(1 + 𝑥 )𝑥 (5)lim(1 − 3𝑥 )𝑥 (6)lim(1 + 𝑡𝑎𝑛2 𝑥 )𝑐𝑜𝑡
𝑥→0
𝑥→0
𝑥→0
(7) lim (𝑥[ln(𝑥 + 1) − 𝑙𝑛𝑥])
𝑥→∞
Know him well to avoid hell)‫(جهنم‬
Accept advice to gain paradise)‫(الجنة‬
58
2𝑥
Extreme values )‫(القيم القصوى‬of functions:
Definition)‫(تعريف‬:
Let 𝑓(𝑥 ) be a function with domain D. Then 𝑓(𝑥 ) has an absolute
maximum)‫ (عظمى‬value on D at a point c if 𝑓(𝑥 ) ≤ 𝑓(𝑐 ) ∀ 𝑥𝜖𝐷.
And absolute minimum)‫ (صغرى‬value on D at c if 𝑓(𝑥 ) ≥
𝑓(𝑐 ) ∀ 𝑥𝜖𝐷
Inflection point)‫(نقطة االنقالب‬:
An inflection point is a point on the graph at which function is
continuous and at which the graph change from concave up to
concave down or vice Vera.
Critical points)‫(النقاط الحرجه‬:
An interior)‫ (الداخليه‬point of the domain of a function 𝑓(𝑥 ) where
𝑓̀(𝑥 ) is zero or undefined is critical point of 𝑓(𝑥 )
First derivative test )‫(اختبار‬for extreme:
Suppose that c is a critical point of a continuous function 𝑓(𝑥 ), and
that 𝑓(𝑥 ) is differentiable at every point in some interval
containing c except possibly at c itself. Moving across c from left to
right:
(1)
If 𝑓̀(𝑥 ) changes from positive to negative at c, then 𝑓(𝑥 )
has a local maximum at c.
(2)
If 𝑓̀(𝑥 ) changes from negative to positive at c, then 𝑓(𝑥 )
has a local minimum at c.
59
(3)
If 𝑓̀(𝑥 ) dos not change sign at c(that 𝑓̀(𝑥 ) positive on
both sides of c or negative on both sides) then 𝑓(𝑥 ) has no
local extreme at c.
Example:
Find the critical values and determine the extreme value if:
(1)𝑓(𝑥 ) = 𝑥 2 − 4𝑥 + 6
(2 ) 𝑓 (𝑥 ) = 𝑥 4
Solution:
(1)𝑓(𝑥 ) = 𝑥 2 − 4𝑥 + 6 ⟹ 𝑓 ̀= 2𝑥 − 4 = 0 ⟹ 𝑥 = 2
𝑓̀(2 − 𝜖 ) = 2(2 − 𝜖 ) − 4 = −2𝜖
𝑓̀(2 + 𝜖 ) = 2(2 + 𝜖 ) − 4 = 2𝜖
Since the sign change from – to + then 𝑓(𝑥 ) has minimum value at
𝑥=2
----------------------------------------------------------------------------------------(2)𝑓(𝑥 ) = 𝑥 4 ⟹ 𝑓 ̀= 4𝑥 3 = 0 ⟹ 𝑥 = 0
𝑓̀(0 − 𝜖 ) = 4(0 − 𝜖 )3 = −4𝜖 3
𝑓̀(0 + 𝜖 ) = 4(0 + 𝜖 )3 = 4𝜖 3
Since the sign change from – to + then 𝑓(𝑥 ) has minimum value at
𝑥=0
60
Second derivative test for extreme:
Suppose 𝑓̀ (𝑥 ) is continuous on an open interval that contains 𝑥 =
0
1. If 𝑓̀(𝑐 ) = 0 𝑎𝑛𝑑 𝑓̀ (𝑐 ) < 0, then 𝑓(𝑥 )has local maximum at
𝑥=𝑐
2. If 𝑓̀(𝑐 ) = 0 𝑎𝑛𝑑 𝑓̀ (𝑐 ) > 0, then 𝑓(𝑥 )has local minimum at
𝑥=𝑐
3. If 𝑓̀(𝑐 ) = 0 𝑎𝑛𝑑 𝑓̀ (𝑐 ) = 0, then the test fails. The function
𝑓(𝑥 ) may have a local maximum, a local minimum, or neither.
Example:
(1)𝑦 = 𝑥 3 − 5𝑥 2 − 8𝑥
(2) 𝑦 = 𝑥 4 − 4𝑥 3 + 4𝑥 2 + 2
Solution:
(1)𝑦 = 𝑥 3 − 5𝑥 2 − 8𝑥
⟹
𝑑𝑦
= 3𝑥 2 − 10𝑥 − 8 = 0
𝑑𝑥
⟹ (3𝑥 + 2)(𝑥 − 4) = 0 ⟹ 𝑥 =
−2
𝑎𝑛𝑑 𝑥 = 4
3
𝑑2 𝑦
⟹ 2 = 6𝑥 − 10
𝑑𝑥
𝑑2 𝑦
−2
⟹ 2|
= 6 ( ) − 10 = −14 < 0
𝑑𝑥 𝑥=−2
3
3
So 𝑥 =
−2
3
is maximum point
61
𝑑2 𝑦
⟹ 2|
= 6(4) − 10 = 14 > 0
𝑑𝑥 𝑥=4
So 𝑥 = 4 is minimum point
----------------------------------------------------------------------------------------(2) 𝑦 = 𝑥 4 − 4𝑥 3 + 4𝑥 2 + 2
⟹
𝑑𝑦
= 4𝑥 3 − 12𝑥 2 + 8𝑥 = 0
𝑑𝑥
⟹ 4𝑥 (𝑥 2 − 3𝑥 + 2) = 0 ⟹ 𝑥 = 0 𝑎𝑛𝑑 𝑥 2 − 3𝑥 + 2 = 0
⟹ (𝑥 − 2)(𝑥 − 1) = 0
⟹ 𝑥 = 2 𝑎𝑛𝑑 𝑥 = 1
𝑑2 𝑦
⟹ 2 = 12𝑥 2 − 24𝑥 + 8
𝑑𝑥
𝑑2 𝑦
⟹ 2|
=0+0+8=8>0
𝑑𝑥 𝑥=0
So 𝑥 = 0 is minimum point
𝑑2 𝑦
⟹ 2|
= 12(2) − 24(2) + 8 = −16 < 0
𝑑𝑥 𝑥=2
So 𝑥 = 2 is maximum point
𝑑2 𝑦
⟹ 2|
= 12(1) − 24(1) + 8 = −4 < 0
𝑑𝑥 𝑥=1
So 𝑥 = 2 is maximum point
62
Exercise:
(A)
(1)𝑓(𝑥 ) = 𝑥 2 + 8𝑥 + 6 (2)𝑓(𝑥 ) = 𝑥 4 + 1 (3) 𝑓(𝑥 ) = 𝑥 4 − 2𝑥 2
(4)𝑓(𝑥 ) = 𝑥 3 + 3𝑥 − 2 (5)𝑓(𝑥 ) = 𝑥 3 − 8 (6) 𝑓(𝑥 ) = 𝑥 2 +
1
1
(7)𝑓(𝑥 ) = 𝑥 3 − 𝑥 2 − 2𝑥 + 1
3
2
(9) 𝑦 = 𝑥 2 + 2𝑥 − 3
(8)𝑓(𝑥 ) = 𝑥 2 − 12𝑥 + 20
(10) 𝑦 = (2 − 𝑥 )3
(10) 𝑦 = 𝑥 3 + 2𝑥 2 − 4𝑥 − 8
(11) 𝑦 = 𝑥 3 +
(12) 𝑦 = (𝑥 − 4)4 (𝑥 + 3)3
(B)
(1) 𝑦 = 15𝑥 3 − 𝑥 5
(2) 𝑦 = −𝑥 4 + 4𝑥 3 − 4𝑥 + 1
1
(3 ) 𝑦 = 𝑥 4 − 𝑥 6
2
(4) 𝑦 = 𝑥 4 − 8𝑥 2 + 16
(5) 𝑦 = 𝑥 √8 −
𝑥3
(6 ) 𝑦 = 2
3𝑥 + 1
𝑥2
(7) 𝑦 = −𝑥 2 − 3𝑥 + 3
(8) 𝑦 = 𝑥 3 + 3𝑥 2 + 3𝑥 + 1
(9) 𝑦 = 15𝑥 3 − 𝑥 5
(10) 𝑦 = 𝑥 2 √5 − 𝑥
1
(11)𝑦 = 𝑥 3 (𝑥 2 − 4)
432
𝑥2
(12) 𝑦 = 2𝑥 3 − 18𝑥
63
48
𝑥
Partial derivative ‫االشتقاق الجزئي‬
Let 𝐴 = 𝐴(𝑥, 𝑦, 𝑧) be a function of more than one variable then
the partial derivative of 𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑥is written as.
𝜕𝐴
𝐴(𝑥 + ∆𝑥, 𝑦, 𝑧) − 𝐴(𝑥, 𝑦, 𝑧)
= 𝐴𝑥 = lim
∆𝑥→0
𝜕𝑥
∆𝑥
And the partial derivative of 𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑦is
𝜕𝐴
𝐴(𝑥, 𝑦 + ∆𝑦, 𝑧) − 𝐴(𝑥, 𝑦, 𝑧)
= 𝐴𝑦 = lim
∆𝑦→0
𝜕𝑦
∆𝑦
And also the partial derivative of 𝐴 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑧 is
𝜕𝐴
𝐴(𝑥, 𝑦, 𝑧 + ∆𝑧) − 𝐴(𝑥, 𝑦, 𝑧)
= 𝐴𝑧 = lim
∆𝑧→0
𝜕𝑧
∆𝑧
Example:
Find all of the first order partial derivatives for the following
functions.
1) 𝐴(𝑥, 𝑦, 𝑧) = 𝑥 2 𝑦 + 𝑥𝑦𝑧 − 𝑦𝑧 2
2) 𝐴(𝑥, 𝑦, 𝑧) = 𝑦𝑐𝑜𝑠𝑥𝑦 − 𝑥𝑒 𝑦𝑧 + 𝑥 3 𝑦 2
64
Solution:
1)
𝜕𝐴
= 𝐴𝑥 = 2𝑥𝑦 + 𝑦𝑧 − 0
𝜕𝑥
𝜕𝐴
= 𝐴𝑦 = 𝑥 2 + 𝑥𝑧 − 𝑧 2
𝜕𝑦
𝜕𝐴
= 𝐴𝑧 = 0 + 𝑥𝑦 − 2𝑦𝑧
𝜕𝑧
---------------------------------------------------------------------------------------------
2)
𝜕𝐴
= 𝐴𝑥 = −𝑦 2 𝑠𝑖𝑛𝑥𝑦 − 𝑒 𝑦𝑧 + 3𝑥 2 𝑦 2
𝜕𝑥
𝜕𝐴
= 𝐴𝑦 = (𝑦) ∙ (−𝑥𝑠𝑖𝑛𝑥𝑦) + (𝑐𝑜𝑠𝑥𝑦) ∙ (1) − 𝑥𝑧𝑒 𝑦𝑧 + 2𝑥 3 𝑦
𝜕𝑦
= 𝑐𝑜𝑠𝑥𝑦 − 𝑥𝑦𝑠𝑖𝑛𝑥𝑦 − 𝑥𝑧𝑒 𝑦𝑧 + 2𝑥 3 𝑦
𝜕𝐴
= 𝐴𝑧 = 0 + 𝑥𝑦𝑒 𝑦𝑧 − 0
𝜕𝑧
Example:
Find 𝑓𝑥 , 𝑓𝑦 and 𝑓𝑧 if 𝑓 = tan−1 𝑥𝑦
65
Solution:
𝑓𝑥 =
𝑦
1 + (𝑥𝑦)2
,
𝑓𝑦 =
𝑥
𝑎𝑛𝑑
1 + (𝑥𝑦)2
𝑓𝑧 = 𝑧𝑒𝑟𝑜
Example:
Find 𝑧𝑥 , 𝑧𝑦 𝑖𝑓 (1) 𝑧 =
𝑥2
𝑦
+
𝑦2
𝑥
(2) 𝑧 = sin(2𝑥 + 3𝑦)
Solution:
2𝑥 𝑦 2
(1) 𝑧𝑥 =
−
𝑦 𝑥2
,
−𝑥 2 2𝑦
𝑧𝑦 = 2 +
𝑦
𝑥
(2)𝑧𝑥 = 2 sin(2𝑥 + 3𝑦) ,
𝑧𝑦 = 3sin(2𝑥 + 3𝑦)
Higher partial derivatives: ‫المشتقات الجزئية العليا‬
The second derivative of a function 𝐴(𝑥, 𝑦, 𝑧) is found by
derivatives of the first one as:
𝜕𝐴
𝜕( 𝜕𝑥 )
𝜕𝑥
𝜕2𝐴
= 𝜕𝑥 2 And also written as 𝐴𝑥𝑥
66
𝜕𝐴
𝜕( 𝜕𝑥 )
𝜕𝑦
𝜕2𝐴
= 𝜕𝑦𝜕𝑥 And also written as 𝐴𝑦𝑥
𝜕𝐴
𝜕( 𝜕𝑥 )
𝜕𝑧
=
𝜕2𝐴
𝜕𝑧𝜕𝑥
And also written as 𝐴𝑧𝑥
----------------------------------------------------------------------------------------𝜕𝐴
)
𝜕𝑦
𝜕(
𝜕𝑥
𝜕2𝐴
= 𝜕𝑥𝜕𝑦 And also written as 𝐴𝑥𝑦
𝜕𝐴
𝜕(𝜕𝑦 )
𝜕𝑦
=
𝜕𝐴
𝜕(𝜕𝑦 )
𝜕𝑧
=
𝜕2𝐴
𝜕𝑦 2
𝜕2𝐴
𝜕𝑧𝜕𝑦
And also written as 𝐴𝑦𝑦
And also written as 𝐴𝑧𝑦
----------------------------------------------------------------------------------------𝜕𝐴
𝜕( 𝜕𝑧 )
𝜕𝑥
𝜕2𝐴
= 𝜕𝑥𝜕𝑧 And also written as 𝐴𝑥𝑧
𝜕𝐴
𝜕( 𝜕𝑧 )
𝜕𝑦
𝜕2𝐴
= 𝜕𝑦𝜕𝑧 And also written as 𝐴𝑧𝑦
𝜕𝐴
𝜕( 𝜕𝑧 )
𝜕𝑧
=
𝜕2𝐴
𝜕𝑧 2
And also written as 𝐴𝑧𝑧
But 𝐴𝑥𝑦 = 𝐴𝑦𝑥 , 𝐴𝑥𝑧 = 𝐴𝑧𝑥 and 𝐴𝑥𝑦 = 𝐴𝑧𝑦
Also we can find upper derivatives as the same way.
67
Example:
Prove that𝐴𝑥𝑦 = 𝐴𝑦𝑥 ,𝐴𝑥𝑧 = 𝐴𝑧𝑥 and
𝐴𝑥𝑦 = 𝐴𝑧𝑦 , 𝑖𝑓 𝐴 = 𝑥𝑦𝑧 −
𝑥𝑐𝑜𝑠𝑦
Solution:
𝐴𝑥 = 𝑦𝑧 − 𝑐𝑜𝑠𝑦
, 𝐴𝑥𝑦 = 𝑧 + 𝑠𝑖𝑛𝑦
𝐴𝑦 = 𝑥𝑧 + 𝑥𝑠𝑖𝑛𝑦 ,
𝐴𝑧 = 𝑥𝑦 ,
, 𝐴𝑥𝑧 = 𝑦
𝐴𝑦𝑥 = 𝑧 + 𝑠𝑖𝑛𝑦 ,
𝐴𝑧𝑦 = 𝑥
𝐴𝑦𝑧 = 𝑥 , 𝐴𝑧𝑥 = 𝑦
∴ 𝐴𝑥𝑦 = 𝐴𝑦𝑥 = 𝑧 + 𝑠𝑖𝑛𝑦
∴ 𝐴𝑥𝑧 = 𝐴𝑧𝑥 = 𝑦
∴ 𝐴𝑦𝑧 = 𝐴𝑧𝑦 = 𝑥
Example:
Find all the second order derivatives for
(1)𝑓 = 𝑐𝑜𝑠2𝑥 − 𝑥 2 𝑒 5𝑦 + 3𝑦 2
(2)𝐴 = 𝑥 2 + 𝑦 2 + 𝑒 𝑥𝑦 +
sinh(𝑦𝑧)
Solution:
(1) 𝑓𝑥 = −2𝑠𝑖𝑛2𝑥 − 2𝑥𝑒 5𝑦
𝑓𝑦 = −5𝑥 2 𝑒 5𝑦 + 6𝑦
68
𝑓𝑥𝑥 = −4𝑐𝑜𝑠2𝑥 − 2𝑒 5𝑦 𝑓𝑥𝑦 = −10𝑥𝑒 5𝑦 𝑓𝑦𝑦 = −25𝑥 2 𝑒 5𝑦 + 6
(2)
𝐴𝑥 = 2𝑥 + 𝑦𝑒 𝑥𝑦 + 0
𝐴𝑥𝑥 = 2 + 𝑦 2 𝑒 𝑥𝑦 + 0
𝐴𝑦𝑥 = 0 + 𝑥𝑦𝑒 𝑥𝑦 + 𝑒 𝑥𝑦
𝐴𝑧𝑥 = 𝑧𝑒𝑟𝑜
𝐴𝑦 = 2𝑦 + 𝑥𝑒 𝑥𝑦 + 𝑧𝑐𝑜𝑠ℎ(𝑦𝑧)
𝐴𝑥𝑦 = 0 + 𝑥𝑦𝑒 𝑥𝑦 + 𝑒 𝑥𝑦 + 0
𝐴𝑦𝑦 = 2 + 𝑥 2 𝑒 𝑥𝑦 + 𝑧 2 sinh(𝑦𝑧)
𝐴𝑧𝑦 = 0 + 0 + 𝑧 ∙ (𝑦𝑠𝑖𝑛ℎ(𝑦𝑧)) + 𝑐𝑜𝑠ℎ(𝑦𝑧)
𝐴𝑧 = 0 + 0 + 0 + 𝑦𝑐𝑜𝑠ℎ(𝑦𝑧)
𝐴𝑥𝑧 = 𝑧𝑒𝑟𝑜
𝐴𝑦𝑧 = 𝑦 ∙ (𝑧𝑠𝑖𝑛ℎ(𝑦𝑧)) + 𝑐𝑜𝑠ℎ(𝑦𝑧)
𝐴𝑧𝑧 = 𝑦 2 sinh(𝑦𝑧)
69
Total derivative:‫التفاضل الكلي‬
If 𝐴 = 𝐴(𝑥, 𝑦, 𝑧) is continuous function then
𝑑𝐴 =
𝜕𝐴
𝜕𝐴
𝜕𝐴
𝑑𝑥 +
𝑑𝑦 +
𝑑𝑧
𝜕𝑥
𝜕𝑦
𝜕𝑧
Is called the total derivative of a function 𝐴(𝑥, 𝑦, 𝑧)
Example:
Find the total derivative to
(1)𝐴(𝑥, 𝑦, 𝑧) = sin
−1
2𝑥 + ln(𝑦 + 𝑧
2)
𝑡3𝑟6
(2 ) 𝑢 = 2
𝑠
Solution:
(1 ) 𝐴 𝑥 =
Since
2
√1 − 4𝑥 2
𝜕𝐴
1
2𝑧
, 𝐴𝑦 =
, 𝐴𝑧 =
(𝑦 + 𝑧 2 )
(𝑦 + 𝑧 2 )
𝜕𝐴
𝜕𝐴
𝑑𝐴 = 𝜕𝑥 𝑑𝑥 + 𝜕𝑦 𝑑𝑦 + 𝜕𝑧 𝑑𝑧
then:
2
1
2𝑧
𝑑𝐴 = (
) 𝑑𝑥 + (
)
𝑑𝑦
+
(
) 𝑑𝑧
(𝑦 + 𝑧 2 )
(𝑦 + 𝑧 2 )
√1 − 4𝑥 2
----------------------------------------------------------------------------------------3𝑡 2 𝑟 6
(2) 𝑢𝑡 =
𝑠2
6𝑡 3 𝑟 5
, 𝑢𝑟 =
𝑠2
−2𝑡 3 𝑟 6
, 𝑢𝑠 =
𝑠3
70
3𝑡 2 𝑟 6
6𝑡 3 𝑟 5
2𝑡 3 𝑟 6
𝑑𝑢 =
𝑑𝑡 +
𝑑𝑟 − 3 𝑑𝑠
𝑠2
𝑠2
𝑠
PROBLEMS)‫(مسائل‬:
(A)Find the total derivative for the following functions.
(1 ) 𝐴 = 𝑧 2 +
𝑦
𝑥2 + 𝑦2
(3) 𝐻 = √𝑒 𝑥+2𝑧 − 𝑦 2
(2) 𝐵 = 𝑧𝑠𝑖𝑛𝑥𝑦 + 3𝑒 2𝑧
(4) 𝑀 = 𝑒 𝑥 + 2𝑒 3𝑦 − 𝑧 + 𝑒 𝑧
(5) 𝐴 = sin−1 (𝑥 + 2𝑦) + 𝑙𝑛𝑧 2 (6) G = xy 2 + yz 2 + xyz
2
(7 ) L = (x + y
3
2 )2
𝑦
(8) 𝐴 = 𝑙𝑛√𝑥 2 + 𝑦 2
(9) 𝑁 = tan−1 ( )
(10) 𝐴 = √9 − 𝑥 2 − 𝑦 2
(11) 𝑧 = (𝑥 2 + 𝑦 2 )
(12) 𝐹 = 2 2
𝑢 +𝑣
𝑧
(13) 𝐺 =
𝑥𝑠𝑖𝑛𝑦
𝑧2
9𝑢
(14) 𝐻 = √𝑥 2 + 𝑙𝑛(3𝑥 + 5𝑦 2 )
(15) 𝐴 = 𝑥 3 𝑧 2 − 5𝑥𝑦 5 𝑧 + 𝑥 2 − 𝑦 3
(B) Find all of the first order partial derivatives for the following
functions.
(1) 𝑧 = 2𝑥 2 − 3𝑥𝑦 + 4𝑦 2 + 𝑥 2 + 𝑥𝑦
(2 ) 𝐴 = 𝑒 𝑥
2 +𝑥𝑦
(3) 𝑒 𝑥 + 2𝑒 3𝑦 + 𝑒 𝑧 = 𝑊 (4) 𝑧 = sin−1 (𝑥 + 2𝑦) + 𝑙𝑛𝑦 2
(5) xy 2 + yz 2 + xyz = F
3
(6 ) z = (x 2 + y 2 )2
71
(7 ) x 2 + y 2 + z 2 = G
(8) z = tan−1 (x 2 y) + tan−1 (xy 2 )
(9) M = x 2 (2y + 3z) + y 2 (3x − 4z) + z 2 (x − 2y) − xyz
(10) xy + yz + zx = U
(11) z = 2x 2 − 3xy + 4y 2
(C)Prove that:
(1) 𝑧𝑥𝑥 + 𝑧𝑦𝑦 = 0 𝑖𝑓 𝑧 = 𝑒 𝑥 𝑐𝑜𝑠𝑦
(2)𝑧𝑥𝑥 + 𝑧𝑦𝑦 = 0
𝑖𝑓 𝑧 = 𝑥 2 − 𝑦 2
Chain Rule)‫(قاعدة السلسه‬
The functions to be considered in the following will be having
continuous first partial derivatives then if:
(1) 𝑧 = 𝑓(𝑥, 𝑦) 𝑎𝑛𝑑 𝑥 = 𝑔(𝑡) , 𝑦 = ℎ(𝑡)
𝑡ℎ𝑒𝑛:
𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
=
∙
+
∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡
Example:
𝑑𝑧
Find 𝑑𝑡 𝑖𝑓 𝑧 = 𝑥 2 𝑐𝑜𝑠𝑦 𝑎𝑛𝑑 𝑥 = 3𝑡 2 + 1 , 𝑦 = 𝑐𝑜𝑠5𝑡
Solution:
𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦
=
∙
+
∙
𝑑𝑡 𝜕𝑥 𝑑𝑡 𝜕𝑦 𝑑𝑡
𝑧𝑥 = 2𝑥𝑐𝑜𝑠𝑦 ,
𝑧𝑦 = −𝑥 2 𝑠𝑖𝑛𝑦 ,
72
𝑑𝑥
= 6𝑡 ,
𝑑𝑡
𝑑𝑦
= −5𝑠𝑖𝑛5𝑡
𝑑𝑡
𝑑𝑧
= (2𝑥𝑐𝑜𝑠𝑦)(6𝑡) + (−𝑥 2 𝑠𝑖𝑛𝑦)(−5𝑠𝑖𝑛5𝑡)
𝑑𝑡
(2) 𝑧 = 𝑓(𝑥, 𝑦) 𝑎𝑛𝑑 𝑥 = 𝑔(𝑢, 𝑣) , 𝑦 = ℎ(𝑢, 𝑣)𝑡ℎ𝑒𝑛:
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
∙
+
∙
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢
∂z ∂z ∂x ∂z ∂y
=
∙ + ∙
∂v ∂x ∂v ∂y ∂v
Example:
𝐹𝑖𝑛𝑑 𝑧𝑢 𝑎𝑛𝑑 𝑧𝑣 𝑖𝑓:
𝑧 = 𝑥 2 + 3𝑥𝑦 + 𝑦 2 𝑎𝑛𝑑 𝑥 = 𝑒 4𝑢 𝑐𝑜𝑠𝑣 , 𝑦 = 𝑒 7𝑢 𝑠𝑖𝑛𝑣
Solution:
(𝑖)
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
∙
+
∙
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢
𝑧𝑥 = 2𝑥 + 3𝑦 , 𝑧𝑦 = 3𝑥 + 2𝑦 , 𝑥𝑢 = 4𝑒 4𝑢 𝑐𝑜𝑠𝑣 , 𝑦𝑢 = 7𝑒 7𝑢 𝑠𝑖𝑛𝑣
𝜕𝑧
= (2𝑥 + 3𝑦 )(4𝑒 4𝑢 𝑐𝑜𝑠𝑣) + (3𝑥 + 2𝑦)(7𝑒 7𝑢 𝑠𝑖𝑛𝑣)
𝜕𝑢
(𝑖𝑖)
𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑧 𝜕𝑦
=
∙
+
∙
𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣
𝑥𝑣 = −𝑒 4𝑢 𝑠𝑖𝑛𝑣 ,
𝑦𝑣 = 𝑒 7𝑢 𝑐𝑜𝑠𝑣
𝜕𝑧
= (2𝑥 + 3𝑦 )(−𝑒 4𝑢 𝑠𝑠𝑖𝑛𝑣) + (3𝑥 + 2𝑦)(𝑒 7𝑢 𝑐𝑜𝑠𝑣)
𝜕𝑣
73
(3) 𝑧 = 𝑓(𝑥, 𝑦) 𝑎𝑛𝑑 𝑦 = ℎ(𝑥 ) 𝑡ℎ𝑒𝑛:
𝑑𝑧 𝜕𝑧 𝑑𝑥 𝜕𝑧 𝑑𝑦 𝜕𝑧 𝜕𝑧 𝑑𝑦
=
∙
+
∙
=
+
∙
𝑑𝑥 𝜕𝑥 𝑑𝑥 𝜕𝑦 𝑑𝑥 𝜕𝑥 𝜕𝑦 𝑑𝑥
Example:
Compute for 𝑧 = 𝑙𝑛𝑥𝑦 + 𝑦 3
Solution:
, 𝑦 = cos(𝑥 2 + 1)
𝑑𝑧 𝜕𝑧 𝜕𝑧 𝑑𝑦
=
+
∙
𝑑𝑥 𝜕𝑥 𝜕𝑦 𝑑𝑥
𝜕𝑧
𝑦
=
𝜕𝑥 𝑥𝑦
𝜕𝑧
𝑥
=
+ 3𝑦 2 ,
𝜕𝑦 𝑥𝑦
,
𝑑𝑦
= −2𝑥𝑠𝑖𝑛(𝑥 2 + 1)
𝑑𝑥
𝑑𝑧
𝑦
𝑥
𝑥
=
+ ( + 3𝑦 2 ) ( + 3𝑦 2 )
𝑑𝑥 𝑥𝑦
𝑥𝑦
𝑥𝑦
Exercise:
𝑑𝑧
(A)Compute 𝑑𝑡 for each of the following:
(1) 𝑧 = 𝑥𝑒 𝑥𝑦 , 𝑥 = 𝑡 2 ,
𝑦 = 𝑡 −1
(2) 𝑧 = 𝑥 2 𝑦 2 + 𝑦𝑐𝑜𝑠𝑥 ,
𝑥 = 𝑙𝑛𝑡 2 , 𝑦 = 𝑠𝑖𝑛4𝑡
(3) 𝑧 = 𝑒 𝑥𝑦 𝑐𝑜𝑠𝑥 , 𝑥 = 𝑐𝑜𝑠𝑡 , 𝑦 = √𝑡 2 + 1
(4) 𝑧 = 𝑒 𝑥𝑦
2
𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑡𝑐𝑜𝑠𝑡 , 𝑦 = 𝑡𝑠𝑖𝑛𝑡
(5) 𝑧 = 𝑥𝑠𝑖𝑛𝑦 − 𝑦𝑐𝑜𝑠𝑥
(6 ) 𝑧 = 𝑒 𝑥
2 +𝑥𝑦
𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑒 3𝑡 , 𝑦 = 𝑐𝑜𝑠𝑡
𝑤ℎ𝑒𝑟𝑒 𝑥 = tan(3𝑡) , 𝑦 = 25𝑡
(7) 𝑧 = 𝑒 𝑥 + 2𝑒 3𝑦 𝑤ℎ𝑒𝑟𝑒 𝑥 = (𝑡 2 + 1)3 , 𝑦 = √𝑡
74
(8) 𝑧 = sin−1 (𝑥 + 2𝑦) + 𝑙𝑛𝑦 2 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑐𝑜𝑠ℎ𝑡 , 𝑦 = sin−1 2𝑡
2
(9 ) z = (x + y
(B)Compute
𝑑𝑧
𝑑𝑥
3
2 )2
where x = sin(2t) ,
y = lnt 2
for each of the following:
(1) 𝑧 = 𝑥𝑐𝑜𝑠3𝑦 + 𝑥 3 𝑦 5 − 𝑒 𝑥𝑦
𝑤ℎ𝑒𝑟𝑒 𝑦 = 3𝑥 + 𝑙𝑛𝑥 2
(2) 𝑧 = 2𝑥 2 − 3𝑥𝑦 + 4𝑦 2 + 𝑥 2 + 𝑥𝑦 𝑤ℎ𝑒𝑟𝑒 𝑦 = 𝑐𝑜𝑠2𝑥
(3 ) 𝑧 = 𝑒 𝑥
2 +3𝑦
𝑤ℎ𝑒𝑟𝑒 𝑦 = cos −1 3𝑥
(4) 𝑧 = 𝑥 2 + 𝑥𝑦 + 𝑦 2
𝑤ℎ𝑒𝑟𝑒
(5) 𝑧 = 𝑥𝑠𝑖𝑛ℎ𝑦 + 𝑦𝑐𝑜𝑠𝑥
(C)Compute
𝑑𝑧
𝑑𝑦
𝑥𝑦
𝑦 = 𝑒 6𝑥
𝑤ℎ𝑒𝑟𝑒 𝑦 = 𝑡𝑎𝑛𝑥 2
for each of the following:
(1) 𝑧 = 𝑥𝑒 + 𝑠𝑖𝑛𝑦 𝑤ℎ𝑒𝑟𝑒 𝑥 = 35𝑦
(2) 𝑧 = 𝑥 2 𝑦 2 − 𝑡𝑎𝑛𝑥 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑙𝑛(𝑦 + 3)
(3) 𝑧 = 𝑒 𝑥𝑦 + 𝑐𝑜𝑠(𝑦𝑥 ) 𝑤ℎ𝑒𝑟𝑒 𝑥 = √𝑦 + 2
2
2
(4) 𝑧 = 𝑒 (𝑥 −𝑦 )
𝑤ℎ𝑒𝑟𝑒 𝑥 = √𝑦 + 2
−1
(5 ) 𝑧 = 𝑥 (𝑥 2 + 𝑦 2 ) 2
𝜕𝑧
𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑒 𝑦
2
𝜕𝑧
(D)Compute
𝑎𝑛𝑑
for each of the following:
𝜕∅
𝜕𝑟
(1) 𝑧 = 𝑥 3 − 𝑥𝑦 + 𝑦 3 𝑖𝑓 𝑥 = 𝑟𝑐𝑜𝑠∅ , 𝑦 = 𝑟𝑠𝑖𝑛∅
(2) 𝑧 = 𝑠𝑖𝑛
𝑦
𝑥
𝑖𝑓 𝑥 = 𝑟 2 + 2∅ , 𝑦 = 4𝑟 − 2∅3
𝑦
(3) 𝑧 = tan−1 ( )
𝑥
𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑙𝑛𝑟 , 𝑦 = 𝑒 3∅
75
‫∅‪𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑐𝑜𝑠𝑟 2 , 𝑦 = 𝑡𝑎𝑛2‬‬
‫𝑦 ‪(4) 𝑧 = 𝑥𝑦 3 + 2𝑥 3‬‬
‫∅‪(5) 𝑧 = 𝑒 𝑥𝑦 + cos(𝑥 + 𝑦 2 ) 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑒 𝑟 − 𝑒 3∅ , 𝑦 = 𝑒 𝑟 − 𝑒 3‬‬
‫‪(E)Prove that:‬‬
‫‪1‬‬
‫𝑧‬
‫𝑓𝑖‬
‫𝑦𝑦 ‪𝛼 2‬‬
‫= 𝑥𝑥𝑧 )‪(1‬‬
‫𝑦𝛼 ‪𝑧 = 𝑓 (𝑢) + 𝑔(𝑣) 𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑥 + 𝛼𝑦 𝑎𝑛𝑑 𝑣 = 𝑢 −‬‬
‫𝑓𝑖‬
‫‪1‬‬
‫‪(𝑢𝑠 )2‬‬
‫‪2‬‬
‫𝑠‬
‫‪2‬‬
‫‪(2) (𝑢𝑥 )2 − (𝑢𝑦 ) = (𝑢𝑟 )2 −‬‬
‫𝑠‪𝑥 = 𝑟𝑠𝑖𝑛ℎ𝑠 , 𝑦 = 𝑟𝑐𝑜𝑠ℎ‬‬
‫𝑓𝑖‬
‫‪1‬‬
‫‪(𝑢𝜃 )2‬‬
‫‪2‬‬
‫𝑟‬
‫‪2‬‬
‫‪(3) (𝑢𝑥 )2 + (𝑢𝑦 ) = (𝑢𝑟 )2 −‬‬
‫𝜃𝑠𝑜𝑐𝑟 = 𝑦 ‪𝑥 = 𝑟𝑠𝑖𝑛𝜃 ,‬‬
‫وال ترج السماحة من بخيل‬
‫ورزقك ليس ينقصه التأني‬
‫وال حزن يدوم وال سرور‬
‫إذا ما آنت ذا قلب قنوع‬
‫‪𝑢 = 𝑓(𝑥, 𝑦) ,‬‬
‫‪𝑢 = 𝑓(𝑥, 𝑦) ,‬‬
‫فما في النار للظمآن ماء‬
‫وليس يزيد في الرزق العناء‬
‫وال بؤس عليك وال رخاء‬
‫فأنت ومالك الدنيا سواء‬
‫‪Some properties:‬‬
‫‪(1)Trigonometric functions‬‬
‫‪76‬‬
(A)
(1) 𝑠𝑖𝑛(𝑥 ± 𝑦) = 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 ± 𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑦
(2) cos(𝑥 ± 𝑦) = 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 ∓ 𝑠𝑖𝑛𝑥𝑠𝑖𝑛𝑦
(3) tan(𝑥 ± 𝑦) =
𝑡𝑎𝑛𝑥 ± 𝑡𝑎𝑛𝑦
1 ∓ 𝑡𝑎𝑛𝑥𝑡𝑎𝑛𝑦
(B)
1
(1) 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 = [sin(𝑥 − 𝑦) + 𝑠𝑖𝑛(𝑥 + 𝑦)]
2
1
(2) 𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑦 = [cos(𝑥 − 𝑦) − cos(𝑥 + 𝑦)]
2
1
(3) 𝑐𝑜𝑠𝑥𝑐𝑜𝑠𝑦 = [cos(𝑥 − 𝑦) + cos(𝑥 + 𝑦)]
2
(C)
1
(1) 𝑠𝑖𝑛2 𝑥 = [1 − cos2x]
2
1
(2) 𝑐𝑜𝑠 2 𝑥 = [1 + cos2x]
2
(3) 𝑠𝑖𝑛2𝑥 = 2𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥
(4) 𝑐𝑜𝑠2𝑥 = 2𝑐𝑜𝑠 2 𝑥 − 1 = 𝑐𝑜𝑠 2 𝑥 − 𝑠𝑖𝑛2 𝑥
(5) 𝑡𝑎𝑛2𝑥 =
2𝑡𝑎𝑛𝑥
1 − 𝑡𝑎𝑛2 𝑥
(D)
77
(1) 𝑠𝑖𝑛2 𝑥 + 𝑐𝑜𝑠 2 𝑥 = 1
(2) 1 + 𝑡𝑎𝑛2 𝑥 = 𝑠𝑒𝑐 2 𝑥
(3) 1 + 𝑐𝑜𝑡 2 𝑥 = 𝑐𝑒𝑠𝑒𝑐 2 𝑥
(E)
𝑥+𝑦
𝑥−𝑦
(1) 𝑠𝑖𝑛𝑥 + 𝑠𝑖𝑛𝑦 = 2 sin (
) cos (
)
2
2
(2) 𝑐𝑜𝑠𝑥 − 𝑐𝑜𝑠𝑦 = −2sin(
𝑥+𝑦
𝑥−𝑦
)sin(
)
2
2
(F)
(1) sin(𝑥 ± 2𝜋) = 𝑠𝑖𝑛𝑥
(2) cos(𝑥 ± 2𝜋) = 𝑐𝑜𝑠𝑥
𝜋
(3) sin ( − 𝑥) = 𝑐𝑜𝑠𝑥
2
𝜋
(4)cos ( − 𝑥) = 𝑠𝑖𝑛𝑥
2
(2)Hyperbolic functions:
(A)
(1) sinh(x ± y) = 𝑠𝑖𝑛ℎ𝑥𝑐𝑜𝑠ℎ𝑦 ± 𝑐𝑜𝑠ℎ𝑥𝑠𝑖𝑛ℎ𝑦
(2) cosh(𝑥 ± 𝑦) = 𝑐𝑜𝑠ℎ𝑥𝑐𝑜𝑠ℎ𝑦 ± 𝑠𝑖𝑛ℎ𝑥𝑠𝑖𝑛ℎ𝑦
(3) 𝑠𝑖𝑛ℎ𝑥 + 𝑠𝑖𝑛ℎ𝑦 = 2sinh(
𝑥+𝑦
𝑥−𝑦
)cosh(
)
2
2
(B)
78
‫‪(1) 𝑐𝑜𝑠ℎ2 𝑥 − 𝑠𝑖𝑛ℎ2 𝑥 = 1‬‬
‫‪1‬‬
‫]𝑥‪(2) 𝑠𝑖𝑛ℎ2 𝑥 = [−1 + 𝑐𝑜𝑠ℎ2‬‬
‫‪2‬‬
‫‪1‬‬
‫] 𝑥‪(3) 𝑐𝑜𝑠ℎ2 𝑥 = [1 + 𝑐𝑜𝑠ℎ2‬‬
‫‪2‬‬
‫𝑥 ‪(4) 𝑡𝑎𝑛ℎ2 𝑥 = 1 − 𝑠𝑒𝑐ℎ2‬‬
‫𝑥 ‪(5) 𝑐𝑜𝑡ℎ2 𝑥 = 1 + 𝑐𝑜𝑠𝑒𝑐ℎ2‬‬
‫اعلل النفس باالمال ارقبها‬
‫مااضيق العيش لوال فسحة االمل‬
‫‪Without hope the heart would break‬‬
‫‪79‬‬