MA-123: Calculus

1st Semester

Lecture -7

Dr. Muhammad Shabbir

Assistant Professor
University of Engineering and Technology
Lahore Pakistan

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 Derivatives of Transcendental Functions
Transcendental Function: A family of non-algebraic functions
is called transcendental functions.
Types of Transcendental Functions:
 Trigonometric Functions
 Inverse Trigonometric Functions
 Hyperbolic Functions
 Inverse Hyperbolic Functions
 Exponential Functions
 Logarithmic Functions

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 Derivatives of Trigonometric Functions:
𝑑
1) (sin 𝑥 ) = cos 𝑥
𝑑𝑥
𝑑
2) (cos 𝑥) = − sin 𝑥
𝑑𝑥
𝑑
3) (tan 𝑥) = sec 2 𝑥
𝑑𝑥
𝑑
4) (cot 𝑥) = − csc 2 𝑥
𝑑𝑥
𝑑
5) (sec 𝑥) = sec 𝑥 tan 𝑥
𝑑𝑥
𝑑
6) (csc 𝑥) = − csc 𝑥 cot 𝑥
𝑑𝑥
𝑑𝑦
Example 1: Find if
𝑑𝑥
a) y = 𝑥 2 + sin 𝑥
b) y = 𝑥 2 sin 𝑥
sin 𝑥
c) y =
𝑥

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Solution:
a) y = 𝑥 2 + sin 𝑥
𝑑𝑦
= 2𝑥 + cos 𝑥
𝑑𝑥
b) y = 𝑥 2 sin 𝑥
𝑑𝑦
= 𝑥 2 cos 𝑥 + 2𝑥 sin 𝑥
𝑑𝑥
sin 𝑥
c) y=
𝑥
𝑑𝑦 𝑥 cos 𝑥 − sin 𝑥
=
𝑑𝑥 𝑥2
𝑑2𝑦
Example 2: Find 2 if y = sec 𝑥 ?
𝑑𝑥
Solution: y = sec 𝑥
𝑑𝑦
= sec 𝑥 tan 𝑥
𝑑𝑥
𝑑2𝑦
= sec 𝑥 sec 2 𝑥 + tan 𝑥 sec 𝑥 tan 𝑥
𝑑𝑥 2

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 𝑑2𝑦
= sec 3 𝑥 + sec 𝑥 tan2 𝑥
𝑑𝑥 2

Derivatives of Inverse Trigonometric Functions:

𝑑 −1 1
1) sin 𝑥 = 2
𝑑𝑥 √1−𝑥
𝑑 −1
2) cos −1 𝑥 =
𝑑𝑥 √1−𝑥 2
𝑑 1
3) tan−1 𝑥 =
𝑑𝑥 1+𝑥 2
𝑑 −1
4) cot −1 𝑥 =
𝑑𝑥 1+𝑥 2
𝑑 1
5) sec −1 𝑥 =
𝑑𝑥 |𝑥|√𝑥 2 −1
𝑑 −1
6) csc −1 𝑥 =
𝑑𝑥 |𝑥|√𝑥 2 −1

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Derivatives of Hyperbolic Functions:
𝑑
1) sinh 𝑥 = cosh 𝑥
𝑑𝑥
𝑑
2) cosh 𝑥 = sinh 𝑥
𝑑𝑥
𝑑
3) tanh 𝑥 = sech2 𝑥
𝑑𝑥
𝑑
4) coth 𝑥 = −csch2 𝑥
𝑑𝑥
𝑑
5) sech 𝑥 = − sech 𝑥 tanh 𝑥
𝑑𝑥
𝑑
6) csch 𝑥 = −coth 𝑥 csch 𝑥
𝑑𝑥
Derivatives of Inverse Hyperbolic Functions:
𝑑 −1 1
1) sinh 𝑥 = 2
𝑑𝑥 √𝑥 +1
𝑑 −1 1
2) cosh 𝑥=
𝑑𝑥 √𝑥 2 −1

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 𝑑 −1 1
3) tanh 𝑥= |𝑥 | < 1
𝑑𝑥 1−𝑥 2
𝑑 1
4) coth−1 𝑥 = |𝑥 | > 1
𝑑𝑥 1−𝑥 2
𝑑 −1 −1
5) sech 𝑥= 0<x<1
𝑑𝑥 𝑥 √1−𝑥 2
𝑑𝑦 −1 −1
6) csch 𝑥= x>0
𝑑𝑥 𝑥 √1+𝑥 2

Derivatives of Exponential and Logarithmic Functions:

𝑑 𝑥
1) 𝑎 = 𝑎 𝑥 ln 𝑎
𝑑𝑥
𝑑 1
2) log 𝑎 𝑥 =
𝑑𝑥 𝑥 ln 𝑎
𝑑
3) 𝑒𝑥 = 𝑒𝑥
𝑑𝑥
𝑑 1
4) ln 𝑥 =
𝑑𝑥 𝑥

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Note: In any of the above formulas if the argument is other than
x then first treats it as x and then multiply its derivative.
e.g.
1) y=𝑒 5𝑥
𝑑𝑦
= 5𝑒 5𝑥
𝑑𝑥
2) y=sin−1 5𝑥
𝑑𝑦 1 𝑑
= (5𝑥)
𝑑𝑥 √1 − (5𝑥 )2 𝑑𝑥
1 5
= .5 =
√1−25𝑥 2 √1−25𝑥 2
Example 3:
1) y=𝑒 𝑎𝑥 sin 𝑏𝑥
𝑑𝑦 𝑎𝑥 𝑑 𝑑
=𝑒 (sin 𝑏𝑥) + sin 𝑏𝑥 (𝑒 𝑎𝑥 )
𝑑𝑥 𝑑𝑥 𝑑𝑥
=𝑒 cos 𝑏𝑥 (𝑏) + sin 𝑏𝑥 𝑒 𝑎𝑥 (𝑎)
𝑎𝑥

=𝑒 𝑎𝑥 (𝑏 cos 𝑏𝑥 + 𝑎 sin 𝑏𝑥 )
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2) If 𝑓(𝑥 ) = 𝑒 𝑥 (1 + ln 𝑥) , then find 𝑓 ′ (𝑥 )
′( ) 𝑥 𝑑 ( ) ( )
𝑑 𝑥
𝑓 𝑥 =𝑒 1 + ln 𝑥 + 1 + ln 𝑥 𝑒
𝑑𝑥 𝑑𝑥
1
=𝑒 𝑥 ( ) + (1+ln x) 𝑒𝑥
𝑥
𝑥 1
=𝑒 ( + 1 + ln 𝑥)
𝑥
𝑑𝑦
3) If 𝑦 = (sin 2∅ − cos 3∅)2 , then find
𝑑∅
𝑑𝑦 𝑑
= 2(sin 2∅ − cos 3∅) (sin 2∅ − cos 3∅)
𝑑∅ 𝑑∅
=2(sin 2∅ − cos 3∅)(2cos 2∅ + 3sin 3∅)
=2(sin 2∅ − cos 3∅)(2cos 2∅ + 3sin 3∅)

ln 𝑥 𝑑𝑦
4) If 𝑦 = , then find
𝑥 𝑑𝑥
𝑑 𝑑
𝑥 𝑑𝑥(ln 𝑥)−ln 𝑥 𝑑𝑥(𝑥)
=
𝑥2
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 1
𝑥 −ln 𝑥.1
𝑥
=
𝑥2
𝑑𝑦 1 − ln 𝑥
=
𝑑𝑥 𝑥2

Exercise 3.5
Q (1-34) (Odd)
Exercise 7.3
Q (7-24, 27-32) (Odd)
Exercise 7.6
Q (21-42) (Odd)
Exercise 7.7
Q (13-22, 25-36) (Odd)

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