general differentiation differentiation of sine & cosine name ......sec 2.5 – general...
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Sec 2.5 – General Differentiation Differentiation of Sine & Cosine Name:
(1) Let’s try to prove the derivative of the function 풇(풙) = 풔풊풏(풙) using the definition of the derivative.
풇′(풙) = 퐥퐢퐦풉→ퟎ
풔풊풏(풙+ 풉) − 풔풊풏(풙)
풉
We could use the sum and difference trig identities to substitute 풔풊풏(풙+ 풉) = 풔풊풏(풙)풄풐풔(풉) + 풄풐풔(풙)풔풊풏(풉)
풇′(풙) = 퐥퐢퐦풉→ퟎ
풔풊풏(풙)풄풐풔(풉) + 풄풐풔(풙)풔풊풏(풉) − 풔풊풏(풙)
풉
We can rearrange the terms of the numerator
풇′(풙) = 퐥퐢퐦풉→ퟎ
풔풊풏(풙)풄풐풔(풉) − 풔풊풏(풙) + 풄풐풔(풙)풔풊풏(풉)
풉
We could then factor out 풔풊풏(풙) from the first two terms.
풇′(풙) = 퐥퐢퐦풉→ퟎ
풔풊풏(풙)(풄풐풔(풉) − ퟏ) + 풄풐풔(풙)풔풊풏(풉)
풉
Next, we could use limit laws to rewrite the following statement:
풇 (풙) = 풔풊풏(풙) ∙ 퐥퐢퐦풉→ퟎ
(풄풐풔(풉) − ퟏ)
풉+ 풄풐풔(풙) ∙ 퐥퐢퐦
풉→ퟎ풔풊풏(풉)
풉
(A) Which leaves us with two indeterminate form limits. We will need to use the Squeeze Limit Theorem. Let’s first investigate the limit lim → . Consider using the following diagram for which 0 ≤ ℎ < .
푨풓풄푳풆풏품풕풉푩푫 ≤ 푩푪
풉 ≤ 퐭퐚퐧풉
풉 ≤ 퐬퐢퐧 풉퐜퐨퐬 풉
풉 ∙ 퐜퐨퐬 풉 ≤ 퐬퐢퐧 풉
퐜퐨퐬 풉 ≤ 퐬퐢퐧 풉풉
퐴푟푐퐿푒푛푔푡ℎ = 푟 ∙ 휃
퐿푒푛푔푡ℎ퐵퐷 = 1 ∙ ℎ
퐿푒푛푔푡ℎ퐵퐷 = ℎ tanℎ =
tanℎ = 퐵퐶
tan휃 =
푫푬 ≤ 푨풓풄푳풆풏품풕풉푩푫
퐬퐢퐧 풉 ≤ 풉
퐬퐢퐧 풉풉 ≤ 풉풉
퐬퐢퐧 풉풉 ≤ ퟏ
퐴푟푐퐿푒푛푔푡ℎ = 푟 ∙ 휃
퐿푒푛푔푡ℎ퐵퐷 = 1 ∙ ℎ
퐿푒푛푔푡ℎ퐵퐷 = ℎ sinℎ =
sinℎ = 퐷퐸
sin휃 =
퐜퐨퐬 풉 ≤ 퐬퐢퐧 풉풉≤ ퟏ
퐥퐢퐦풉→ퟎ
퐜퐨퐬 풉 ≤ 퐥퐢퐦풉→ퟎ
퐬퐢퐧 풉풉≤ 퐥퐢퐦
풉→ퟎퟏ
ퟏ ≤ 퐥퐢퐦풉→ퟎ
퐬퐢퐧 풉풉≤ ퟏ
퐥퐢퐦풉→ퟎ
퐬퐢퐧 풉풉 = ퟏ
By the Squeeze Limit Theorem:
퐥퐢퐦풉→ퟎ
퐬퐢퐧 풉풉
= 퐥퐢퐦풉→ퟎ
퐬퐢퐧( 풉)풉
sin(−푥) = − sin(푥)
Although we were only working with a right-handed limit we could find the left hand limit by substituting h with – h.
Since the function 푓(푥) = sin(푥) is an odd function
If we make the appropriate substitution:
퐥퐢퐦풉→ퟎ 퐬퐢퐧( 풉)풉 = 퐥퐢퐦풉→ퟎ 퐬퐢퐧(풉)
풉 =퐥퐢퐦풉→ퟎ 퐬퐢퐧(풉)
풉 = ퟏ
M. Winking © Unit 2-5 page 37
(B) Next, let’s investigate the limit 퐥퐢퐦풉→ퟎ퐜퐨퐬(풉) ퟏ
풉 and consider multiplying it by something equivalent to 1.
퐥퐢퐦풉→ퟎ
(풄풐풔(풉) − ퟏ)
풉 ∙
(풄풐풔(풉) + ퟏ)(풄풐풔(풉) + ퟏ)
After expanding the numerator, we would have.
= 퐥퐢퐦풉→ퟎ
풄풐풔ퟐ(풉) − ퟏ
풉 ∙ (풄풐풔(풉) + ퟏ)
Then, we could use the trigonometric Pythagorean identity sin (ℎ) + cos (ℎ) = 1 rearranged to cos (ℎ) − 1 = −sin (ℎ)
= 퐥퐢퐦풉→ퟎ
−풔풊풏ퟐ(풉)
풉 ∙ (풄풐풔(풉) + ퟏ)
Next, we could separate the fraction into 2 pieces using limit laws:
= 퐥퐢퐦풉→ퟎ
풔풊풏(풉)풉
∙ 퐥퐢퐦풉→ퟎ
−풔풊풏(풉)
(풄풐풔(풉) + ퟏ)
The limit on the left we just determined was lim → = 1 and the limit on the right we could use direct substitution:
lim→
( ) = ퟏ ∙−풔풊풏(ퟎ)
(풄풐풔(ퟎ) + ퟏ) = ퟏ ∙ퟎ
(ퟏ+ ퟏ) = ퟎ
Finally, we can go back to the original limit with the two indeterminate limits lim →( ) = 0 and lim → = 1:
풇 (풙) = 풔풊풏(풙) ∙ 퐥퐢퐦풉→ퟎ
(풄풐풔(풉) − ퟏ)
풉+ 풄풐풔(풙) ∙ 퐥퐢퐦
풉→ퟎ풔풊풏(풉)
풉
풇 (풙) = 풔풊풏(풙) ∙ 0 + 풄풐풔(풙) ∙ 1
풇 (풙) = 풄풐풔(풙)
(2) Let’s try to prove the derivative of the function 품(풙) = 풄풐풔(풙) using the definition of the derivative.
품′(풙) = 퐥퐢퐦풉→ퟎ
풄풐풔(풙 + 풉) − 풄풐풔(풙)
풉
We could use the sum and difference trig identities to substitute 풄풐풔(풙+ 풉) = 풄풐풔(풙)풄풐풔(풉) − 풔풊풏(풙)풔풊풏(풉)
품′(풙) = 퐥퐢퐦풉→ퟎ
풄풐풔(풙)풄풐풔(풉) − 풔풊풏(풙)풔풊풏(풉)− 풄풐풔(풙)
풉
We can rearrange the terms of the numerator
품′(풙) = 퐥퐢퐦풉→ퟎ
풄풐풔(풙)풄풐풔(풉) − 풄풐풔(풙) − 풔풊풏(풙)풔풊풏(풉)
풉
We could then factor out 풄풐풔(풙) from the first two terms.
품′(풙) = 퐥퐢퐦풉→ퟎ
풄풐풔(풙)(풄풐풔(풉) − ퟏ) − 풔풊풏(풙)풔풊풏(풉)
풉
Next, we could use limit laws to rewrite the following statement:
품 (풙) = 풄풐풔(풙) ∙ 퐥퐢퐦풉→ퟎ
(풄풐풔(풉) − ퟏ)
풉− 풔풊풏(풙) ∙ 퐥퐢퐦
풉→ퟎ풔풊풏(풉)
풉
Finally, we can go back to the original limit with the two indeterminate limits lim →( ) = 0 and lim → = 1:
품 (풙) = 풄풐풔(풙) ∙ 퐥퐢퐦풉→ퟎ
(풄풐풔(풉) − ퟏ)
풉− 풔풊풏(풙) ∙ 퐥퐢퐦
풉→ퟎ풔풊풏(풉)
풉= 풄풐풔(풙) ∙ ퟎ − 풔풊풏(풙) ∙ ퟏ = −풔풊풏(풙)
M. Winking © Unit 2-5 page 38
1. Using the various methods shown determine the general derivative of the following:
A. 푓(푥) = 3 sin 푥 + 푥 B. 푓(푥) = 푒 ∙ cos(푥)
C. 푦 = 푥 ∙ cos(푥) D. 푦 = ( )
E. 푓(푥) = cos(푥) ∙ sin(푥) F. 푦 =( )
Power Rule Given: 푓(푥) = 푎 ∙ 푥
,where a and n are constants.
푓′(푥) = 푛 ∙ 푎 ∙ 푥
Exponential Rule Given: 푓(푥) = 푎
,where a is a constant.
푓′(푥) = 푙푛(푎) ∙ 푎
Natural Exponent Given: 푓(푥) = 푒 ,where e ≈2.7182818.
푓′(푥) = 푒
Product Rule
,where f and g are differentiable.
(푓 ∙ 푔)′ = 푓(푥) ∙ 푔′(푥) + 푔(푥) ∙ 푓′(푥) Quotient Rule
,where f and g are differentiable.
푓푔 =
푔(푥)푓 (푥) − 푓(푥)푔 (푥)
푔(푥)
Derivative of Sine Given: 푓(푥) = sin(푥)
푓′(푥) = cos(푥)
Derivative of Cosine Given: 푓(푥) = cos(푥)
푓′(푥) = −sin(푥)
M. Winking © Unit 2-5 page 39
2. Given 푓(푥) = 푥 ∙ 푐표푠(푥), determine the exact value of 푓′ .
3. Given 푓(푥) =( ), determine
the exact value of 푓′ .
4. Given 푓(푥) = 푔(푥) ∙ 푠푖푛(푥) , 푔 = 3 , and 푔′ = −2, determine the exact value of 푓′ .
5. Given 푓(푥) = ( )( ) , 푔 = 3 , and
푔′ = −2, determine the exact value of 푓′ .
6. Given 푓(푥) = 푞(푥) ∙ 푠푖푛(푥) , determine the exact value of 푓′(0).
M. Winking © Unit 2-5 page 40
7. Determine the higher order derivatives
a. sin sinf x x x b. Find 3
3 sind xdx
Find f x .
8. Find the equation of the line tangent to 푓(푥) = 푥 + sin 푥 at 푥 = .
9. Find the equation of the line tangent to 푓(푥) = 푔(푥) ∙ 푐표푠(푥) at 푥 = given:
푔 = 4 푔′ = 2
M. Winking © Unit 2-5 page 41