general differentiation differentiation of sine & cosine name ......sec 2.5 – general...

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)


풔풊풏(풙)풄풐풔(풉) + 풄풐풔(풙)풔풊풏(풉) − 풔풊풏(풙)

풉

We can rearrange the terms of the numerator


풔풊풏(풙)풄풐풔(풉) − 풔풊풏(풙) + 풄풐풔(풙)풔풊풏(풉)

풉

We could then factor out 풔풊풏(풙) from the first two terms.


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


풉 ∙

(풄풐풔(풉) + ퟏ)(풄풐풔(풉) + ퟏ)

After expanding the numerator, we would have.


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)


−풔풊풏ퟐ(풉)

풉 ∙ (풄풐풔(풉) + ퟏ)

Next, we could separate the fraction into 2 pieces using limit laws:


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


풉+ 풄풐풔(풙) ∙ 퐥퐢퐦


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)


풄풐풔(풙)풄풐풔(풉) − 풔풊풏(풙)풔풊풏(풉)− 풄풐풔(풙)

풉

We can rearrange the terms of the numerator


풄풐풔(풙)풄풐풔(풉) − 풄풐풔(풙) − 풔풊풏(풙)풔풊풏(풉)

풉

We could then factor out 풄풐풔(풙) from the first two terms.


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


풉− 풔풊풏(풙) ∙ 퐥퐢퐦


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


풉− 풔풊풏(풙) ∙ 퐥퐢퐦


풉= 풄풐풔(풙) ∙ ퟎ − 풔풊풏(풙) ∙ ퟏ = −풔풊풏(풙)


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)


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).


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


general differentiation differentiation of sine & cosine name ......sec 2.5 – general...

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