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After defining the limit and calculating a few, we introduced the limit laws. Today we do the same for the derivative. We calculate a few and introduce laws which allow us to computer more. The Power Rule shows us how to compute derivatives of polynomials, and we can also find directly the derivative of sine and cosine.
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. . . . . .
Section2.3BasicDifferentiationRules
V63.0121.034, CalculusI
September24, 2009
Announcements
I Quiznextweek(uptoSection2.1)I Seewebsiteforup-to-dateevents.
. . . . . .
Outline
Recall
DerivativessofarDerivativesofpowerfunctionsbyhandThePowerRule
DerivativesofpolynomialsThePowerRuleforwholenumberpowersThePowerRuleforconstantsTheSumRuleTheConstantMultipleRule
Derivativesofsineandcosine
Derivative. . . . . .
. . . . . .
Recall: thederivative
DefinitionLet f beafunctionand a apointinthedomainof f. Ifthelimit
f′(a) = limh→0
f(a + h) − f(a)h
= limx→a
f(x) − f(a)x− a
exists, thefunctionissaidtobe differentiableat a and f′(a) isthederivativeof f at a.Thederivative…
I …measurestheslopeofthelinethrough (a, f(a)) tangenttothecurve y = f(x);
I …representstheinstantaneousrateofchangeof f at aI …producesthebestpossiblelinearapproximationto f near
a.
. . . . . .
Notation
Newtoniannotation Leibniziannotation
f′(x) y′(x) y′dydx
ddx
f(x)dfdx
. . . . . .
Linkbetweenthenotations
f′(x) = lim∆x→0
f(x + ∆x) − f(x)∆x
= lim∆x→0
∆y∆x
=dydx
I Leibnizthoughtofdydx
asaquotientof“infinitesimals”
I Wethinkofdydx
asrepresentinga limit of(finite)difference
quotients, notasanactualfractionitself.I Thenotationsuggeststhingswhicharetrueeventhoughtheydon’tfollowfromthenotation perse
. . . . . .
Outline
Recall
DerivativessofarDerivativesofpowerfunctionsbyhandThePowerRule
DerivativesofpolynomialsThePowerRuleforwholenumberpowersThePowerRuleforconstantsTheSumRuleTheConstantMultipleRule
Derivativesofsineandcosine
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h= lim
h→0
2x��h + h�2
��h= lim
h→0(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h= lim
h→0
2x��h + h�2
��h= lim
h→0(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h= lim
h→0
2x��h + h�2
��h= lim
h→0(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h
= limh→0
2x��h + h�2
��h= lim
h→0(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h= lim
h→0
2x��h + h�2
��h
= limh→0
(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Derivativeofthesquaringfunction
ExampleSuppose f(x) = x2. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x2
h= lim
h→0
2x��h + h�2
��h= lim
h→0(2x + h) = 2x.
So f′(x) = 2x.
. . . . . .
Thesecondderivative
If f isafunction, sois f′, andwecanseekitsderivative.
f′′ = (f′)′
Itmeasurestherateofchangeoftherateofchange!
Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
. . . . . .
Thesecondderivative
If f isafunction, sois f′, andwecanseekitsderivative.
f′′ = (f′)′
Itmeasurestherateofchangeoftherateofchange! Leibniziannotation:
d2ydx2
d2
dx2f(x)
d2fdx2
. . . . . .
Thesquaringfunctionanditsderivatives
. .x
.y
.f
.f′.f′′ I f increasing =⇒ f′ ≥ 0
I f decreasing =⇒ f′ ≤ 0I horizontaltangentat 0
=⇒ f′(0) = 0
. . . . . .
Thesquaringfunctionanditsderivatives
. .x
.y
.f
.f′.f′′ I f increasing =⇒ f′ ≥ 0
I f decreasing =⇒ f′ ≤ 0I horizontaltangentat 0
=⇒ f′(0) = 0
. . . . . .
Thesquaringfunctionanditsderivatives
. .x
.y
.f
.f′
.f′′
I f increasing =⇒ f′ ≥ 0I f decreasing =⇒ f′ ≤ 0I horizontaltangentat 0
=⇒ f′(0) = 0
. . . . . .
Thesquaringfunctionanditsderivatives
. .x
.y
.f
.f′.f′′ I f increasing =⇒ f′ ≥ 0
I f decreasing =⇒ f′ ≤ 0I horizontaltangentat 0
=⇒ f′(0) = 0
. . . . . .
Derivativeofthecubingfunction
ExampleSuppose f(x) = x3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x3
h= lim
h→0
3x2��h + 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f′(x) = 3x2.
. . . . . .
Derivativeofthecubingfunction
ExampleSuppose f(x) = x3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x3
h= lim
h→0
3x2��h + 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f′(x) = 3x2.
. . . . . .
Derivativeofthecubingfunction
ExampleSuppose f(x) = x3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x3
h
= limh→0
3x2��h + 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f′(x) = 3x2.
. . . . . .
Derivativeofthecubingfunction
ExampleSuppose f(x) = x3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x3
h= lim
h→0
3x2��h + 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f′(x) = 3x2.
. . . . . .
Derivativeofthecubingfunction
ExampleSuppose f(x) = x3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x3
h= lim
h→0
3x2��h + 3xh���1
2 + h���2
3
��h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f′(x) = 3x2.
. . . . . .
Thecubingfunctionanditsderivatives
. .x
.y
.f
.f′.f′′I Noticethat f isincreasing, and f′ > 0except f′(0) = 0
I Noticealsothatthetangentlinetothegraphof f at (0, 0) crosses thegraph(contrarytoapopular“definition”ofthetangentline)
. . . . . .
Thecubingfunctionanditsderivatives
. .x
.y
.f
.f′.f′′I Noticethat f isincreasing, and f′ > 0except f′(0) = 0
I Noticealsothatthetangentlinetothegraphof f at (0, 0) crosses thegraph(contrarytoapopular“definition”ofthetangentline)
. . . . . .
Thecubingfunctionanditsderivatives
. .x
.y
.f
.f′
.f′′
I Noticethat f isincreasing, and f′ > 0except f′(0) = 0
I Noticealsothatthetangentlinetothegraphof f at (0, 0) crosses thegraph(contrarytoapopular“definition”ofthetangentline)
. . . . . .
Thecubingfunctionanditsderivatives
. .x
.y
.f
.f′.f′′I Noticethat f isincreasing, and f′ > 0except f′(0) = 0
I Noticealsothatthetangentlinetothegraphof f at (0, 0) crosses thegraph(contrarytoapopular“definition”ofthetangentline)
. . . . . .
Thecubingfunctionanditsderivatives
. .x
.y
.f
.f′.f′′I Noticethat f isincreasing, and f′ > 0except f′(0) = 0
I Noticealsothatthetangentlinetothegraphof f at (0, 0) crosses thegraph(contrarytoapopular“definition”ofthetangentline)
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x) = lim
h→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x) = lim
h→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x) = lim
h→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x)
= limh→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x) = lim
h→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
DerivativeofthesquarerootfunctionExampleSuppose f(x) =
√x = x1/2. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
√x + h−
√x
h
= limh→0
√x + h−
√x
h·√x + h +
√x√
x + h +√x
= limh→0
(�x + h) − �x
h(√
x + h +√x) = lim
h→0
��h
��h(√
x + h +√x)
=1
2√x
So f′(x) =√x = 1
2x−1/2.
. . . . . .
Thesquarerootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0+
f′(x) = ∞ and
f isnotdifferentiableat 0I Noticealso
limx→∞
f′(x) = 0
. . . . . .
Thesquarerootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0+
f′(x) = ∞ and
f isnotdifferentiableat 0I Noticealso
limx→∞
f′(x) = 0
. . . . . .
Thesquarerootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0+
f′(x) = ∞ and
f isnotdifferentiableat 0
I Noticealsolimx→∞
f′(x) = 0
. . . . . .
Thesquarerootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0+
f′(x) = ∞ and
f isnotdifferentiableat 0I Noticealso
limx→∞
f′(x) = 0
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)
= limh→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)
=1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
DerivativeofthecuberootfunctionExampleSuppose f(x) = 3
√x = x1/3. Usethedefinitionofderivativetofind
f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)1/3 − x1/3
h
= limh→0
(x + h)1/3 − x1/3
h· (x + h)2/3 + (x + h)1/3x1/3 + x2/3
(x + h)2/3 + (x + h)1/3x1/3 + x2/3
= limh→0
(�x + h) − �xh
((x + h)2/3 + (x + h)1/3x1/3 + x2/3
)= lim
h→0
��h
��h((x + h)2/3 + (x + h)1/3x1/3 + x2/3
) =1
3x2/3
So f′(x) = 13x
−2/3.
. . . . . .
Thecuberootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0
f′(x) = ∞ and f
isnotdifferentiableat 0I Noticealso
limx→±∞
f′(x) = 0
. . . . . .
Thecuberootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0
f′(x) = ∞ and f
isnotdifferentiableat 0I Noticealso
limx→±∞
f′(x) = 0
. . . . . .
Thecuberootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0
f′(x) = ∞ and f
isnotdifferentiableat 0
I Noticealsolim
x→±∞f′(x) = 0
. . . . . .
Thecuberootfunctionanditsderivatives
. .x
.y
.f
.f′
I Here limx→0
f′(x) = ∞ and f
isnotdifferentiableat 0I Noticealso
limx→±∞
f′(x) = 0
. . . . . .
Onemore
ExampleSuppose f(x) = x2/3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·((x + h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
So f′(x) = 23x
−1/3.
. . . . . .
Onemore
ExampleSuppose f(x) = x2/3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·((x + h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
So f′(x) = 23x
−1/3.
. . . . . .
Onemore
ExampleSuppose f(x) = x2/3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·((x + h)1/3 + x1/3
)
= 13x
−2/3(2x1/3
)= 2
3x−1/3
So f′(x) = 23x
−1/3.
. . . . . .
Onemore
ExampleSuppose f(x) = x2/3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·((x + h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)
= 23x
−1/3
So f′(x) = 23x
−1/3.
. . . . . .
Onemore
ExampleSuppose f(x) = x2/3. Usethedefinitionofderivativetofind f′(x).
Solution
f′(x) = limh→0
f(x + h) − f(x)h
= limh→0
(x + h)2/3 − x2/3
h
= limh→0
(x + h)1/3 − x1/3
h·((x + h)1/3 + x1/3
)= 1
3x−2/3
(2x1/3
)= 2
3x−1/3
So f′(x) = 23x
−1/3.
. . . . . .
Thefunction x 7→ x2/3 anditsderivative
. .x
.y
.f
.f′
I f isnotdifferentiableat0 and lim
x→0±f′(x) = ±∞
I Noticealsolim
x→±∞f′(x) = 0
. . . . . .
Thefunction x 7→ x2/3 anditsderivative
. .x
.y
.f
.f′
I f isnotdifferentiableat0 and lim
x→0±f′(x) = ±∞
I Noticealsolim
x→±∞f′(x) = 0
. . . . . .
Thefunction x 7→ x2/3 anditsderivative
. .x
.y
.f
.f′
I f isnotdifferentiableat0 and lim
x→0±f′(x) = ±∞
I Noticealsolim
x→±∞f′(x) = 0
. . . . . .
Thefunction x 7→ x2/3 anditsderivative
. .x
.y
.f
.f′
I f isnotdifferentiableat0 and lim
x→0±f′(x) = ±∞
I Noticealsolim
x→±∞f′(x) = 0
. . . . . .
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap
y y′
x2 2x1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
Recap: TheTowerofPower
y y′
x2 2x
1
x3 3x2
x1/2 12x
−1/2
x1/3 13x
−2/3
x2/3 23x
−1/3
I Thepowergoesdownbyoneineachderivative
I Thecoefficientinthederivativeisthepoweroftheoriginalfunction
. . . . . .
ThePowerRule
Thereismountingevidencefor
Theorem(ThePowerRule)Let r bearealnumberand f(x) = xr. Then
f′(x) = rxr−1
aslongastheexpressionontheright-handsideisdefined.
I PerhapsthemostfamousruleincalculusI WewillassumeitasoftodayI Wewillproveitmanywaysformanydifferent r.
. . . . . .
TheotherTowerofPower
. . . . . .
Outline
Recall
DerivativessofarDerivativesofpowerfunctionsbyhandThePowerRule
DerivativesofpolynomialsThePowerRuleforwholenumberpowersThePowerRuleforconstantsTheSumRuleTheConstantMultipleRule
Derivativesofsineandcosine
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat.
Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Rememberyouralgebra
FactLet n beapositivewholenumber. Then
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
Proof.Wehave
(x + h)n = (x + h) · (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
=n∑
k=0
ckxkhn−k
Thecoefficientof xn is 1 becausewehavetochoose x fromeachbinomial, andthere’sonlyonewaytodothat. Thecoefficientofxn−1h isthenumberofwayswecanchoose x n− 1 times, whichisthesameasthenumberofdifferent hswecanpick, whichisn.
. . . . . .
Pascal’sTriangle
..1
.1 .1
.1 .2 .1
.1 .3 .3 .1
.1 .4 .6 .4 .1
.1 .5 .10 .10 .5 .1
.1 .6 .15 .20 .15 .6 .1
(x + h)0 = 1
(x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
(x + h)3 = 1x3 + 3x2h + 3xh2 + 1h3
. . . . . .
. . . . . .
Pascal’sTriangle
..1
.1 .1
.1 .2 .1
.1 .3 .3 .1
.1 .4 .6 .4 .1
.1 .5 .10 .10 .5 .1
.1 .6 .15 .20 .15 .6 .1
(x + h)0 = 1
(x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
(x + h)3 = 1x3 + 3x2h + 3xh2 + 1h3
. . . . . .
. . . . . .
Pascal’sTriangle
..1
.1 .1
.1 .2 .1
.1 .3 .3 .1
.1 .4 .6 .4 .1
.1 .5 .10 .10 .5 .1
.1 .6 .15 .20 .15 .6 .1
(x + h)0 = 1
(x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
(x + h)3 = 1x3 + 3x2h + 3xh2 + 1h3
. . . . . .
. . . . . .
Pascal’sTriangle
..1
.1 .1
.1 .2 .1
.1 .3 .3 .1
.1 .4 .6 .4 .1
.1 .5 .10 .10 .5 .1
.1 .6 .15 .20 .15 .6 .1
(x + h)0 = 1
(x + h)1 = 1x + 1h
(x + h)2 = 1x2 + 2xh + 1h2
(x + h)3 = 1x3 + 3x2h + 3xh2 + 1h3
. . . . . .
. . . . . .
Theorem(ThePowerRule)Let r beapositivewholenumber. Then
ddx
xr = rxr−1
Proof.Asweshowedabove,
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
So
(x + h)n − xn
h=
nxn−1h + (stuffwithatleasttwo hsinit)h
= nxn−1 + (stuffwithatleastone h init)
andthistendsto nxn−1 as h → 0.
. . . . . .
Theorem(ThePowerRule)Let r beapositivewholenumber. Then
ddx
xr = rxr−1
Proof.Asweshowedabove,
(x + h)n = xn + nxn−1h + (stuffwithatleasttwo hsinit)
So
(x + h)n − xn
h=
nxn−1h + (stuffwithatleasttwo hsinit)h
= nxn−1 + (stuffwithatleastone h init)
andthistendsto nxn−1 as h → 0.
. . . . . .
ThePowerRuleforconstants
TheoremLet c beaconstant. Then
ddx
c = 0
.
.likeddx
x0 = 0x−1
(although x 7→ 0x−1 isnotdefinedatzero.)
Proof.Let f(x) = c. Then
f(x + h) − f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
. . . . . .
ThePowerRuleforconstants
TheoremLet c beaconstant. Then
ddx
c = 0 .
.likeddx
x0 = 0x−1
(although x 7→ 0x−1 isnotdefinedatzero.)
Proof.Let f(x) = c. Then
f(x + h) − f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
. . . . . .
ThePowerRuleforconstants
TheoremLet c beaconstant. Then
ddx
c = 0 .
.likeddx
x0 = 0x−1
(although x 7→ 0x−1 isnotdefinedatzero.)
Proof.Let f(x) = c. Then
f(x + h) − f(x)h
=c− ch
= 0
So f′(x) = limh→0
0 = 0.
. . . . . .
NewderivativesfromoldThisiswherethecalculusstartstogetreallypowerful!
Calculus
. . . . . .
. . . . . .
RecalltheLimitLaws
FactSuppose lim
x→af(x) = L and lim
x→ag(x) = M and c isaconstant. Then
1. limx→a
[f(x) + g(x)] = L + M
2. limx→a
[f(x) − g(x)] = L−M
3. limx→a
[cf(x)] = cL
4. . . .
. . . . . .
Addingfunctions
Theorem(TheSumRule)Let f and g befunctionsanddefine
(f + g)(x) = f(x) + g(x)
Thenif f and g aredifferentiableat x, thensois f + g and
(f + g)′(x) = f′(x) + g′(x).
Succinctly, (f + g)′ = f′ + g′.
. . . . . .
Proof.Followyournose:
(f + g)′(x) = limh→0
(f + g)(x + h) − (f + g)(x)h
= limh→0
f(x + h) + g(x + h) − [f(x) + g(x)]h
= limh→0
f(x + h) − f(x)h
+ limh→0
g(x + h) − g(x)h
= f′(x) + g′(x)
NotetheuseoftheSumRuleforlimits. Sincethelimitsofthedifferencequotientsforfor f and g exist, thelimitofthesumisthesumofthelimits.
. . . . . .
Proof.Followyournose:
(f + g)′(x) = limh→0
(f + g)(x + h) − (f + g)(x)h
= limh→0
f(x + h) + g(x + h) − [f(x) + g(x)]h
= limh→0
f(x + h) − f(x)h
+ limh→0
g(x + h) − g(x)h
= f′(x) + g′(x)
NotetheuseoftheSumRuleforlimits. Sincethelimitsofthedifferencequotientsforfor f and g exist, thelimitofthesumisthesumofthelimits.
. . . . . .
Proof.Followyournose:
(f + g)′(x) = limh→0
(f + g)(x + h) − (f + g)(x)h
= limh→0
f(x + h) + g(x + h) − [f(x) + g(x)]h
= limh→0
f(x + h) − f(x)h
+ limh→0
g(x + h) − g(x)h
= f′(x) + g′(x)
NotetheuseoftheSumRuleforlimits. Sincethelimitsofthedifferencequotientsforfor f and g exist, thelimitofthesumisthesumofthelimits.
. . . . . .
Proof.Followyournose:
(f + g)′(x) = limh→0
(f + g)(x + h) − (f + g)(x)h
= limh→0
f(x + h) + g(x + h) − [f(x) + g(x)]h
= limh→0
f(x + h) − f(x)h
+ limh→0
g(x + h) − g(x)h
= f′(x) + g′(x)
NotetheuseoftheSumRuleforlimits. Sincethelimitsofthedifferencequotientsforfor f and g exist, thelimitofthesumisthesumofthelimits.
. . . . . .
Scalingfunctions
Theorem(TheConstantMultipleRule)Let f beafunctionand c aconstant. Define
(cf)(x) = cf(x)
Thenif f isdifferentiableat x, sois cf and
(cf)′(x) = c · f′(x)
Succinctly, (cf)′ = cf′.
. . . . . .
Proof.Again, followyournose.
(cf)′(x) = limh→0
(cf)(x + h) − (cf)(x)h
= limh→0
cf(x + h) − cf(x)h
= c limh→0
f(x + h) − f(x)h
= c · f′(x)
. . . . . .
Proof.Again, followyournose.
(cf)′(x) = limh→0
(cf)(x + h) − (cf)(x)h
= limh→0
cf(x + h) − cf(x)h
= c limh→0
f(x + h) − f(x)h
= c · f′(x)
. . . . . .
Proof.Again, followyournose.
(cf)′(x) = limh→0
(cf)(x + h) − (cf)(x)h
= limh→0
cf(x + h) − cf(x)h
= c limh→0
f(x + h) − f(x)h
= c · f′(x)
. . . . . .
Proof.Again, followyournose.
(cf)′(x) = limh→0
(cf)(x + h) − (cf)(x)h
= limh→0
cf(x + h) − cf(x)h
= c limh→0
f(x + h) − f(x)h
= c · f′(x)
. . . . . .
Derivativesofpolynomials
Example
Findddx
(2x3 + x4 − 17x12 + 37
)
Solution
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .
Derivativesofpolynomials
Example
Findddx
(2x3 + x4 − 17x12 + 37
)Solution
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .
Derivativesofpolynomials
Example
Findddx
(2x3 + x4 − 17x12 + 37
)Solution
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .
Derivativesofpolynomials
Example
Findddx
(2x3 + x4 − 17x12 + 37
)Solution
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .
Derivativesofpolynomials
Example
Findddx
(2x3 + x4 − 17x12 + 37
)Solution
ddx
(2x3 + x4 − 17x12 + 37
)=
ddx
(2x3
)+
ddx
x4 +ddx
(−17x12
)+
ddx
(37)
= 2ddx
x3 +ddx
x4 − 17ddx
x12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
. . . . . .
Outline
Recall
DerivativessofarDerivativesofpowerfunctionsbyhandThePowerRule
DerivativesofpolynomialsThePowerRuleforwholenumberpowersThePowerRuleforconstantsTheSumRuleTheConstantMultipleRule
Derivativesofsineandcosine
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1 = cos x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1 = cos x
. . . . . .
AngleadditionformulasSeeAppendixA
.
.
sin(A + B) = sinA cosB + cosA sinB
cos(A + B) = cosA cosB− sinA sinB
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1 = cos x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1 = cos x
. . . . . .
TwoimportanttrigonometriclimitsSeeSection1.4
..θ
.sin θ
.1− cos θ
.θ
.−1 .1
.
.
limθ→0
sin θ
θ= 1
limθ→0
cos θ − 1θ
= 0
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1
= cos x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = ???
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1
= cos x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos x
Proof.Fromthedefinition:
ddx
sin x = limh→0
sin(x + h) − sin xh
= limh→0
( sin x cos h + cos x sin h) − sin xh
= sin x · limh→0
cos h− 1h
+ cos x · limh→0
sin hh
= sin x · 0 + cos x · 1 = cos x
. . . . . .
IllustrationofSineandCosine
. .x
.y
.π .−π2 .0 .π2
.π.sin x
.cos x
I f(x) = sin x hashorizontaltangentswhere f′ = cos(x) iszero.I whathappensatthehorizontaltangentsof cos?
. . . . . .
IllustrationofSineandCosine
. .x
.y
.π .−π2 .0 .π2
.π.sin x.cos x
I f(x) = sin x hashorizontaltangentswhere f′ = cos(x) iszero.
I whathappensatthehorizontaltangentsof cos?
. . . . . .
IllustrationofSineandCosine
. .x
.y
.π .−π2 .0 .π2
.π.sin x.cos x
I f(x) = sin x hashorizontaltangentswhere f′ = cos(x) iszero.I whathappensatthehorizontaltangentsof cos?
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos xddx
cos x = − sin x
Proof.Wealreadydidthefirst. Thesecondissimilar(mutatismutandis):
ddx
cos x = limh→0
cos(x + h) − cos xh
= limh→0
(cos x cos h− sin x sin h) − cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos xddx
cos x = − sin x
Proof.Wealreadydidthefirst. Thesecondissimilar(mutatismutandis):
ddx
cos x = limh→0
cos(x + h) − cos xh
= limh→0
(cos x cos h− sin x sin h) − cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos xddx
cos x = − sin x
Proof.Wealreadydidthefirst. Thesecondissimilar(mutatismutandis):
ddx
cos x = limh→0
cos(x + h) − cos xh
= limh→0
(cos x cos h− sin x sin h) − cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos xddx
cos x = − sin x
Proof.Wealreadydidthefirst. Thesecondissimilar(mutatismutandis):
ddx
cos x = limh→0
cos(x + h) − cos xh
= limh→0
(cos x cos h− sin x sin h) − cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
. . . . . .
DerivativesofSineandCosine
Fact
ddx
sin x = cos xddx
cos x = − sin x
Proof.Wealreadydidthefirst. Thesecondissimilar(mutatismutandis):
ddx
cos x = limh→0
cos(x + h) − cos xh
= limh→0
(cos x cos h− sin x sin h) − cos xh
= cos x · limh→0
cos h− 1h
− sin x · limh→0
sin hh
= cos x · 0− sin x · 1 = − sin x
. . . . . .
Whathavewelearnedtoday?
I ThePowerRule
I ThederivativeofasumisthesumofthederivativesI Thederivativeofaconstantmultipleofafunctionisthatconstantmultipleofthederivative
I ThederivativeofsineiscosineI Thederivativeofcosineistheoppositeofsine.
. . . . . .
Whathavewelearnedtoday?
I ThePowerRuleI ThederivativeofasumisthesumofthederivativesI Thederivativeofaconstantmultipleofafunctionisthatconstantmultipleofthederivative
I ThederivativeofsineiscosineI Thederivativeofcosineistheoppositeofsine.
. . . . . .
Whathavewelearnedtoday?
I ThePowerRuleI ThederivativeofasumisthesumofthederivativesI Thederivativeofaconstantmultipleofafunctionisthatconstantmultipleofthederivative
I ThederivativeofsineiscosineI Thederivativeofcosineistheoppositeofsine.
