Transcendental Functions PDF

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DERIVATIVES OF TRANSCENDENTAL FUNCTIONS TRANSCENDENTAL FUNCTION A function which is not algebraic is called a transcendental function. Functions such as trigonometric, inverse trigonometric, logarithmic, and exponential are classified as transcendental. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 𝐿𝑒𝑡 𝑢 𝑏𝑒 𝑎 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥. 𝑇ℎ𝑒𝑛, 𝑑𝑢

𝐷10 :

𝑑 (sin 𝑢) 𝑑𝑥

= cos 𝑢 ∙

𝐷11 :

𝑑 (cos 𝑢) 𝑑𝑥

= − 𝑠𝑖𝑛 𝑢 ∙

𝐷12 :

𝑑 (tan 𝑢) 𝑑𝑥

= 𝑠𝑒𝑐 2 𝑢 ∙

𝐷13 :

𝑑𝑥

𝐷14 :

𝑑𝑥

𝐷15 :

𝑑𝑥

𝑑𝑥

𝑑𝑢

𝑑𝑥

𝑑𝑢

𝑑𝑥

𝑑

(cot 𝑢) = − 𝑐𝑠𝑐 2 𝑢 ∙

𝑑

(sec 𝑢) = sec 𝑢 ∙ tan 𝑢 ∙

𝑑

(csc 𝑢) = − csc 𝑢 ∙ cot 𝑢 ∙

𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥

ILLUSTRATION: Find the derivative of y with respect to x then simplify the result whenever possible. 1.

𝑦 = cos(𝑥2 + 1) 𝑑𝑢

Let 𝑢 = 𝑥2 + 1 ;

𝑑𝑥

= 2𝑥

∴ 𝑦 = cos 𝑢 By 𝐷11 :

2.

𝑑𝑦

𝑑𝑥

= − sin 𝑢 ∙

𝒅𝒚 𝒅𝒙

= − sin(𝑥2 + 1) ∙ 2𝑥 = −𝟐𝒙 𝐬𝐢𝐧(𝒙𝟐 + 𝟏)

𝑦 = 2 tan

𝑥

2

𝑑𝑢 𝑑𝑥

−𝑥

Let 𝑢 = ∴ 𝑦 = 2 tan 𝑢 − 𝑥

𝑥

2

;

𝑑𝑢

𝑑𝑥

=

1 2

𝑑𝑦 𝑑𝑥

By 𝐷12 :

𝑑𝑢

= 2 𝑠𝑒𝑐 2 𝑢 ∙

𝑑𝑥

−1

1

𝑥

= 2 ∙ 𝑠𝑒𝑐 2 2 ∙ − 1 = 𝑠𝑒𝑐 2

𝑥

2

−1

2

𝟏 + 𝒕𝒂𝒏𝟐 𝜽 = 𝒔𝒆𝒄𝟐 𝜽)

(by the Pythagorean Identity ∴

3.

𝒅𝒚 𝒅𝒙

= 𝒕𝒂𝒏𝟐

𝒙

𝟐

𝑦 = 𝑢2 − 2𝑢 + 4 { 𝑢 = 1 + sec 𝑥 𝑦 = 𝑢2 − 2𝑢 + 4 ;

𝑑𝑦

= 2𝑢 − 2 = 2(𝑢 − 1)

𝑢 = 1 + sec 𝑥

𝑑𝑢

= sec 𝑥 ∙ tan 𝑥

;

𝑑𝑢

𝑑𝑥

By 𝐷7 : 𝑑𝑦

𝑑𝑥

=

𝑑𝑦

𝑑𝑢

𝑑𝑢

∙ 𝑑𝑥

= [2(𝑢 − 1)] ∙ (sec 𝑥 ∙ tan 𝑥) = [2((1 + sec 𝑥) − 1)] ∙ sec 𝑥 ∙ tan 𝑥 ∴

𝒅𝒚 𝒅𝒙

= 𝟐 𝒔𝒆𝒄𝟐 𝒙 𝒕𝒂𝒏 𝒙

4. 𝑦 = 2 cos 𝑥 ∙ sin 2𝑥 − sin 𝑥 ∙ cos 2𝑥 By product rule: 𝑑𝑦

𝑑𝑥 𝒅𝒚

𝒅𝒙

= 2[cos 𝑥 (2 cos 2𝑥) + sin 2𝑥 (− sin 𝑥)] − [sin 𝑥 (−2 sin 2𝑥) + cos 2𝑥 (cos 𝑥)] = 𝟑 𝐜𝐨𝐬 𝒙 𝐜𝐨𝐬 𝟐𝒙

5. 𝑥 = 𝑠𝑒𝑐 2 𝑦 By implicit differentiation: 1 = 2 sec 𝑦 [sec 𝑦 tan 𝑦 𝑑𝑦

=

1 2 𝑠𝑒𝑐 2 𝑦 tan 𝑦

𝒅𝒚

=

𝟏

𝑑𝑥

𝒅𝒙

𝟐

𝑑𝑦

𝑑𝑥

]

𝒄𝒐𝒔𝟐𝒚 𝐜𝐨𝐭 𝒚

2 6. {𝑦 = 𝑢 − 2𝑢 + 4 𝑢 = 1 + sec 𝑥 𝑑𝑦

= 2𝑢 − 2 = 2(𝑢 − 1)

𝑑𝑢

= sec 𝑥 ∙ tan 𝑥

𝑦 = 𝑢 2 − 2𝑢 + 4 ;

𝑑𝑢

𝑢 = 1 + sec 𝑥

𝑑𝑥

;

By 𝐷7 : 𝑑𝑦

𝑑𝑥

=

𝑑𝑦

𝑑𝑢

∙

𝑑𝑢

𝑑𝑥

= [2(𝑢 − 1)] ∙ (sec 𝑥 ∙ tan 𝑥) = [2((1 + sec 𝑥) − 1)] ∙ sec 𝑥 ∙ tan 𝑥 ∴

𝒅𝒚

𝒅𝒙

= 𝟐 𝒔𝒆𝒄𝟐 𝒙 𝒕𝒂𝒏 𝒙

Properties of Inverse Trigonometric Functions If y is a function of x determined by the relation tan y = x , then y is called the inverse tangent function of x and is denoted by 𝒚 = 𝐚𝐫𝐜𝐭𝐚𝐧 𝒙

𝒐𝒓

𝒚 = 𝐭𝐚𝐧−𝟏 𝒙

where the symbols are read as “ the angle whose tangent is x” .

DERIVATIVES: 𝐿𝑒𝑡 𝑢 𝑏𝑒 𝑎 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥. Then, 𝑑

(sin−1 𝑢) =

√1 − 𝑢 2

𝑑

(cos −1 𝑢) =

√1 − 𝑢 2

𝑑

(tan −1 𝑢) =

1+ 𝑢2

𝐷16 :

𝑑𝑥

𝐷17 :

𝑑𝑥

𝐷18 :

𝑑𝑥 𝑑

𝐷19 :

(cot −1 𝑢) = 𝑑𝑥

𝐷20 :

𝑑𝑥

𝐷21 :

𝑑𝑥

1

∙

𝑑𝑢 𝑑𝑥

−1

∙

𝑑𝑢 𝑑𝑥

1

∙

𝑑𝑢 𝑑𝑥

−1

∙

𝑑𝑢 𝑑𝑥

1+ 𝑢 2 1

∙

𝑑𝑢 𝑑𝑥

−1

∙

𝑑𝑢 𝑑𝑥

𝑑

(s𝑒𝑐 −1 𝑢) =

𝑢√ 𝑢 2 −1

𝑑

(csc −1 𝑢) =

𝑢√ 𝑢 2 −1

ILLUSTRATION: 𝑑𝑦

If 𝑦 = cot −1 √𝑥 2 − 2𝑥 , find 𝑑𝑥 in simplest form. Solution: 1

𝑦 = cot −1 √𝑥 2 − 2𝑥 𝐿𝑒𝑡 𝑢 = √𝑥 2 − 2𝑥 = (𝑥 2 − 2𝑥) ⁄ 2 𝑑𝑢

=

𝑑𝑢

=

𝑑𝑥 𝑑𝑥

∴ 𝑦 = cot −1 𝑢

1 (𝑥 2 2 𝑥 −1

− 2𝑥)

√𝑥 2 −2𝑥

−1⁄ 2

[2𝑥 − 2]

By 𝐷19 : 𝑑𝑦

𝑑𝑥

𝑑

𝑑𝑥

−1

=

∙

2 1+ (√𝑥 2 −2𝑥)

1+ 𝑢 2

∙

𝑑𝑢 𝑑𝑥

𝑥 −1 √𝑥 2 −2𝑥

−(𝑥 −1)

=

(𝑥 2 −2𝑥+1)√𝑥 2 −2𝑥

=

− ( 𝑥 −1) (𝑥 −1)2√𝑥 2 −2𝑥

=

−1 (𝑥 −1)√𝑥 2−2𝑥

𝒅𝒚

−1

(cot −1 𝑢) =

= 𝒅𝒙

∙

√𝑥 2 −2𝑥

√𝑥 2 −2𝑥

−√𝒙𝟐 −𝟐𝒙 (𝒙 −𝟏)(𝒙𝟐 −𝟐𝒙)

ILLUSTRATION: 𝑑𝑦

If 𝑦 = (sin−1 4𝑥 )2 , find 𝑑𝑥 in simplest form. Solution: 𝑦 = (sin−1 4𝑥 )2 𝐵𝑦 𝑃𝑜𝑤𝑒𝑟 𝑅𝑢𝑙𝑒: 𝐿𝑒𝑡 𝑢 = sin−1 4𝑥 𝑑𝑢

𝑑𝑥

∴ 𝑦 = 𝑢2 𝑑𝑦 𝑑𝑥

=2𝑢

𝑑𝑢 𝑑𝑥 4 √1 −16𝑥2

= 2 sin−1 4𝑥 ∙

𝒅𝒚

𝒅𝒙

=

8 sin−1 4𝑥 √1 −16𝑥 2

=

𝟖√𝟏 −𝟏𝟔𝒙𝟐 ∙𝐬𝐢𝐧−𝟏 𝟒𝒙

∙

√1 −16𝑥 2 √1 −16𝑥 2

𝟏 −𝟏𝟔 𝒙𝟐

=

1

√1 − (4𝑥)2

∙ [4]

ILLUSTRATION: 𝑦 = tan−1 √𝑥 2 − 1 + csc −1 𝑥 Solution: 𝑑𝑦

= 𝑑𝑥

𝒅𝒚 𝒅𝒙

1

2

1+ (√𝑥 2 −1) 1

𝑥

=

𝑥 2 √𝑥 2 −1

∙

=

𝑥 √𝑥 2 −1

1

+

1

∙ [ ∙ (𝑥 2 − 1) 2

−1⁄ 2

∙ (2𝑥)] +

−1

+

𝑥 √𝑥 2 −1 −1

𝑥 √𝑥 2 −1

=𝟎

EXPONENTIAL AND LOGARITHMIC FUNCTIONS PROPERTIES OF LOGARITHMS 1. log 𝑎 𝑀𝑁 = log 𝑎 𝑀 + 2. log 𝑎

𝑀

𝑁

= log 𝑎 𝑀 −

log 𝑎 𝑁 log 𝑎 𝑁

3. log 𝑎 𝑀𝑝 = 𝑝 ∙ log 𝑎 𝑀 4. log 𝑎 𝑛√𝑀 =

1 𝑛

∙ log 𝑎 𝑀

5. log 𝑎 𝑎 = 1 6. log 𝑎 1 = 0 7. 𝑎 log𝑎 𝑀 = 𝑀 𝒘𝒉𝒆𝒓𝒆 𝒂 > 𝟎 𝒆𝒙𝒄𝒆𝒑𝒕 𝟏 EXPONENTIAL LAWS 1. 𝑎 𝑚 ∙ 𝑎 𝑛 = 𝑎 𝑚+𝑛 2. 𝑎 𝑚 ÷ 𝑎 𝑛 =

𝑎𝑚 𝑎𝑛

3. (𝑎 𝑚 )𝑛 = 𝑎 𝑚𝑛

= 𝑎𝑚 − 𝑛 , 𝑎𝑛 ≠ 0

−1

𝑥 √𝑥 2 −1

∙ (1)

4. (𝑎𝑏)𝑚 = 𝑎 𝑚 ∙ 𝑏𝑚 𝑎 𝑚

5. ( ) 𝑏

𝑎𝑚

=

𝑏𝑚

, 𝑏 ≠0

6. 𝑎 0 = 1 1

7. 𝑎 −𝑛 =

𝑎𝑛

DERIVATIVES: 𝐿𝑒𝑡 𝑢 𝑏𝑒 𝑎 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥. 𝑑

(log 𝑎 𝑢) = log 𝑎 𝑒 ∙

𝑑

(ln 𝑢 ) =

𝐷22 :

𝑑𝑥

𝐷23 :

𝑑𝑥 𝑑 𝑑𝑥

𝐷24 :

𝑑

𝐷25 :

𝑑𝑥

1

𝑢

∙

𝑢

∙

𝑑𝑢

𝑑𝑥

𝑑𝑢 𝑑𝑥

(𝑎 𝑢 ) = 𝑎 𝑢 ∙ ln 𝑎 ∙ (𝑒 𝑢 ) = 𝑒 𝑢 ∙

1

𝑑𝑢

𝑑𝑥

𝑑𝑢 𝑑𝑥

𝒘𝒉𝒆𝒓𝒆 𝒂 > 𝟎 𝒆𝒙𝒄𝒆𝒑𝒕 𝟏

ILLUSTRATION: 𝑑𝑦

If 𝑦 = log 2 √𝑥 2 − 2𝑥 , find 𝑑𝑥 in simplest form. Solution: By 𝐷22 :

𝑑

𝑑𝑥

(log 𝑎 𝑢) = log 𝑎 𝑒 ∙

1

𝑢

∙

𝑑𝑢

𝑑𝑥

𝐿𝑒𝑡 𝑢 = √𝑥 2 − 2𝑥 = (𝑥 2 − 2𝑥 ) 𝑑𝑢

𝑑𝑥

1 (𝑥 2 2

=

𝑑𝑢

𝑑𝑥

=

𝑥 −1 √𝑥 2 −2𝑥

∴ 𝑦 = log 2 𝑢 𝑑𝑦

𝑑𝑥

= log 2 𝑒 ∙

1

𝑢

− 2𝑥)

𝑑𝑢

∙ 𝑑𝑥

−1⁄ 2

[2𝑥 − 2]

1⁄ 2

= log 2 𝑒 ∙ 𝒅𝒚

∴ 𝒅𝒙 = 𝐥𝐨𝐠 𝟐 𝒆 ∙

1

√𝑥 2 −2𝑥

∙

𝑥 −1 √𝑥 2 −2𝑥

𝒙 −𝟏

𝒙𝟐 −𝟐𝒙

ALTERNATIVE SOLUTION: Using a property of logarithm, 𝑦 = log 2 √𝑥 2 − 2𝑥 is equivalent to 1

𝑦= 𝑑𝑦

𝑑𝑥

=

log 2 (𝑥 2 − 2𝑥)

2

1

1

log 2 𝑒 ∙ 𝑥 2 −2𝑥 ∙ (2𝑥 − 2)

2

𝒅𝒚

∴ 𝒅𝒙 = 𝐥𝐨𝐠 𝟐 𝒆 ∙

𝒙 −𝟏 𝒙𝟐 −𝟐𝒙

ILLUSTRATION: If 𝑦 = ln(𝑥 sec 𝑥) , find

𝑑𝑦

𝑑𝑥

in simplest form.

1

∙ 𝑑𝑥

Solution: 𝑑

By 𝐷23 :

(ln 𝑢 ) =

𝑑𝑥

𝑢

𝑑𝑢

𝐿𝑒𝑡 𝑢 = 𝑥 sec 𝑥 𝑑𝑢

𝑑𝑥

=𝑥∙

𝑑

𝑑𝑥

(sec 𝑥) + sec 𝑥 ∙

𝑑

𝑑𝑥

(𝑥 )

= 𝑥 ∙ (sec 𝑥 ∙ tan 𝑥 ) + sec 𝑥 (1) 𝑑𝑢

𝑑𝑥

= 𝑥 sec 𝑥 tan 𝑥 + 𝑠𝑒𝑐 𝑥 ∴ 𝑦 = ln 𝑢

𝑑𝑦

𝑑𝑥

= =

1

∙

𝑢

1

𝑑𝑢 𝑑𝑥

𝑥 sec 𝑥 1

∙ [𝑥 sec 𝑥 tan 𝑥 + 𝑠𝑒𝑐 𝑥]

= 𝑥 sec 𝑥 ∙ [sec 𝑥 (𝑥 tan 𝑥 + 1]

𝒅𝒚

∴

𝒙 𝐭𝐚𝐧 𝒙 + 𝟏

=

𝒅𝒙

𝒙

= 𝒕𝒂𝒏 𝒙 +

𝟏 𝒙

ALTERNATIVE SOLUTION: Using a property of logarithm, 𝑦 = ln(𝑥 sec 𝑥) is equivalent to 𝑦 = ln 𝑥 + ln sec 𝑥 𝑑

By 𝐷23 :

𝒅𝒚

∴ 𝒅𝒙 =

1

(ln 𝑢 ) =

𝑑𝑥

𝑑𝑢

∙ 𝑑𝑥

𝑢

𝑑𝑦 1 1 ∙ (sec 𝑥 ∙ tan 𝑥 ) ∙ (1) = ∙ (1) + 𝑑𝑥 𝑥 sec 𝑥

𝟏

+ 𝐭𝐚𝐧 𝒙

𝒙

ILLUSTRATION: If 𝑦 = 2√𝑥

2

−4

𝑑𝑦

, find 𝑑𝑥 in simplest form.

Solution: 𝑑

By 𝐷24 :

𝑑𝑥

𝑑𝑢

(𝑎 𝑢 ) = 𝑎 𝑢 ∙ ln 𝑎 ∙ 𝑑𝑥 1

𝐿𝑒𝑡 𝑢 = √𝑥 2 − 4 = (𝑥 2 − 4) ⁄2 𝑑𝑢

=

1 2

𝑑𝑢

=

𝑥 1 (𝑥 2 −4) ⁄2

𝑑𝑥

𝑑𝑥

∴

∙ (𝑥 2 − 4)

𝑑𝑦

= 2𝑢 ∙ ln 2 ∙ = 2√𝑥 =

𝒅𝒙

∙ (2𝑥)

𝑦 = 2𝑢 𝑑𝑥

𝒅𝒚

−1⁄ 2

=

2

−4

𝑑𝑢

𝑑𝑥

∙ ln 2 ∙

√ 2 2 𝑥 −4 ∙𝑥 ln 2

√𝑥 2 −4

∙

𝑥

1 (𝑥 2 −4) ⁄2

√𝑥 2 −4

√ 𝟐 𝟐 𝒙 −𝟒 ∙𝒙√𝒙𝟐 −𝟒 ∙𝒍𝒏 𝟐 𝒙𝟐 −𝟒

√𝑥 2 −4

ILLUSTRATION: 2

If 𝑦 = 𝑒 tan(3𝑥 ), find

𝑑𝑦 𝑑𝑥

in simplest form.

Solution: 𝑑

By 𝐷25 :

𝑑𝑥

(𝑒 𝑢 ) = 𝑒 𝑢 ∙

𝑑𝑢

𝑑𝑥

𝐿𝑒𝑡 𝑢 = tan(3𝑥 2 ) 𝑑𝑢

= 𝑠𝑒𝑐 2 (3𝑥 2 ) ∙ [6𝑥]

𝑑𝑢

= 6𝑥 ∙ 𝑠𝑒𝑐 2 (3𝑥 2 )

𝑑𝑥 𝑑𝑥

∴

𝑦 = 𝑒𝑢 𝑑𝑦

𝑑𝑥

= 𝑒𝑢 ∙

𝑑𝑢

𝑑𝑥

= 𝑒 tan(3𝑥 ∴

𝒅𝒚 𝒅𝒙

2)

∙ [6𝑥 ∙ 𝑠𝑒𝑐 2 (3𝑥 2 )]

= 𝟔𝒙 ∙ 𝒔𝒆𝒄𝟐 (𝟑𝒙𝟐) ∙ 𝒆𝒕𝒂𝒏(𝟑𝒙

𝟐)

ILLUSTRATION: 𝑑𝑦

2

If 𝑦 = 𝑥 𝑥 , find 𝑑𝑥 in simplest form. Solution: Taking the natural logarithm of the given equation, 𝑦 = 𝑥𝑥

2 2

ln 𝑦 = ln(𝑥 𝑥 ) Then applying the property of logarithm: ln 𝑦 = 𝑥 2 ∙ ln 𝑥 Using Implicit Differentiation: 1

𝑦

∙

𝑑𝑦 𝑑𝑥

= 𝑥2 ∙

1

𝑥

+ ln 𝑥 ∙ (2𝑥)

∴

𝒅𝒚

𝒅𝒙

= 𝒚 ∙ (𝒙 + 𝟐𝒙 𝐥𝐧 𝒙)...