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Section 3.1Derivatives of Polynomials and Exponentials
Math 1a
February 20, 2008
Announcements
I Problem Sessions Sunday, Thursday, 7pm, SC 310
I ALEKS due today (10% of grade).
I Office hours Wednesday 2/20 2–4pm SC 323
I Midterm I Friday 2/29 in class (up to §3.2)
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Outline
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
![Page 3: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/3.jpg)
Derivative of the squaring function
Example
Suppose f (x) = x2. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x
2
h= lim
h→0
2x�h + h�2
�h
= limh→0
(2x + h) = 2x .
So f ′(x) = 2x.
![Page 4: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/4.jpg)
Derivative of the squaring function
Example
Suppose f (x) = x2. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)2 − x2
h
= limh→0
��x2 + 2xh + h2 −��x
2
h= lim
h→0
2x�h + h�2
�h
= limh→0
(2x + h) = 2x .
So f ′(x) = 2x.
![Page 5: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/5.jpg)
Derivative of the cubing function
Example
Suppose f (x) = x3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x
3
h= lim
h→0
3x2�h + 3xh���
1
2 + h���2
3
�h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f ′(x) = 2x.
![Page 6: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/6.jpg)
Derivative of the cubing function
Example
Suppose f (x) = x3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→0
(x + h)3 − x3
h
= limh→0
��x3 + 3x2h + 3xh2 + h3 −��x
3
h= lim
h→0
3x2�h + 3xh���
1
2 + h���2
3
�h
= limh→0
(3x2 + 3xh + h2
)= 3x2.
So f ′(x) = 2x.
![Page 7: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/7.jpg)
Derivative of the square root function
Example
Suppose f (x) =√
x = x1/2. Use the definition of derivative to findf ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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 = 12x−1/2.
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Derivative of the square root function
Example
Suppose f (x) =√
x = x1/2. Use the definition of derivative to findf ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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 = 12x−1/2.
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Derivative of the cube root function
Example
Suppose f (x) = 3√
x = x1/3. Use the definition of derivative to findf ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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)−�x
h((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.
![Page 10: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/10.jpg)
Derivative of the cube root function
Example
Suppose f (x) = 3√
x = x1/3. Use the definition of derivative to findf ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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)−�x
h((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.
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One more
Example
Suppose f (x) = x2/3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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.
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One more
Example
Suppose f (x) = x2/3. Use the definition of derivative to find f ′(x).
Solution
f ′(x) = limh→0
f (x + h)− f (x)
h= lim
h→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.
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The Power Rule
There is mounting evidence for
Theorem (The Power Rule)
Let r be a real number and f (x) = x r . Then
f ′(x) = rx r−1
as long as the expression on the right-hand side is defined.
I Perhaps the most famous rule in calculus
I We will assume it as of today
I We will prove it many ways for many different r .
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Outline
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
![Page 15: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/15.jpg)
Remember your algebra
FactLet n be a positive whole number. Then
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
Proof.We have
(x + h)n = (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
Each monomial is of the form ckxkhn−k . The coefficient of xn isone because we have to choose x from each binomial, and there’sonly one way to do that. The coefficient of xn−1h is the number ofways we can choose x n − 1 times, which is the same as thenumber of different hs we can pick, which is n.
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Remember your algebra
FactLet n be a positive whole number. Then
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
Proof.We have
(x + h)n = (x + h) · (x + h) · · · (x + h)︸ ︷︷ ︸n copies
Each monomial is of the form ckxkhn−k . The coefficient of xn isone because we have to choose x from each binomial, and there’sonly one way to do that. The coefficient of xn−1h is the number ofways we can choose x n − 1 times, which is the same as thenumber of different hs we can pick, which is n.
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Theorem (The Power Rule)
Let r be a positive whole number. Then
d
dxx r = rx r−1
Proof.As we showed above,
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
So
(x + h)n − xn
h=
nxn−1h + (stuff with at least two hs in it)
h= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h→ 0.
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Theorem (The Power Rule)
Let r be a positive whole number. Then
d
dxx r = rx r−1
Proof.As we showed above,
(x + h)n = xn + nxn−1h + (stuff with at least two hs in it)
So
(x + h)n − xn
h=
nxn−1h + (stuff with at least two hs in it)
h= nxn−1 + (stuff with at least one h in it)
and this tends to nxn−1 as h→ 0.
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The Power Rule for constants
TheoremLet c be a constant. Then
d
dxc = 0
Kind of liked
dxx0 = 0x−1, although x 7→ 0x−1 is not defined at
zero.
Proof.Let f (x) = c . Then
f (x + h)− f (x)
h=
c − c
h= 0
So f ′(x) = limh→0
0 = 0.
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The Power Rule for constants
TheoremLet c be a constant. Then
d
dxc = 0
Kind of liked
dxx0 = 0x−1, although x 7→ 0x−1 is not defined at
zero.
Proof.Let f (x) = c . Then
f (x + h)− f (x)
h=
c − c
h= 0
So f ′(x) = limh→0
0 = 0.
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New derivatives from old
This is where the calculus starts to get really powerful!
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Adding functions
Theorem (The Sum Rule)
Let f and g be functions and define
(f + g)(x) = f (x) + g(x)
Then if f and g are differentiable at x, then so is f + g and
(f + g)′(x) = f ′(x) + g ′(x).
Succinctly, (f + g)′ = f ′ + g ′.
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Proof.Follow your nose:
(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+ lim
h→0
g(x + h)− g(x)
h
= f ′(x) + g ′(x)
Note the use of the Sum Rule for limits. Since the limits of thedifference quotients for for f and g exist, the limit of the sum isthe sum of the limits.
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Scaling functions
Theorem (The Constant Multiple Rule)
Let f be a function and c a constant. Define
(cf )(x) = cf (x)
Then if f is differentiable at x, so is cf and
(cf )′(x) = cf ′(x)
Succinctly, (cf )′ = cf ′.
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Proof.Again, follow your nose.
(cf )′(x) = limhto0
(cf )(x + h)− (cf )(x)
h
= limhto0
cf (x + h)− cf (x)
h
= c limhto0
f (x + h)− f (x)
h
= cf ′(x)
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Derivatives of polynomials
Example
Findd
dx
(2x3 + x4 − 17x12 + 37
)
Solution
d
dx
(2x3 + x4 − 17x12 + 37
)=
d
dx
(2x3)
+d
dxx4 +
d
dx
(−17x12
)+
d
dx(37)
= 2d
dxx3 +
d
dxx4 − 17
d
dxx12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
![Page 27: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/27.jpg)
Derivatives of polynomials
Example
Findd
dx
(2x3 + x4 − 17x12 + 37
)Solution
d
dx
(2x3 + x4 − 17x12 + 37
)=
d
dx
(2x3)
+d
dxx4 +
d
dx
(−17x12
)+
d
dx(37)
= 2d
dxx3 +
d
dxx4 − 17
d
dxx12 + 0
= 2 · 3x2 + 4x3 − 17 · 12x11
= 6x2 + 4x3 − 204x11
![Page 28: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/28.jpg)
Outline
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
![Page 29: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/29.jpg)
Derivative of x 7→ 2x
Example
Let f (x) = 2x . Use a calculator to estimate f ′(0).
SolutionWe have
f ′(0) = limh→0
20+h − 20
h= lim
h→0
2h − 1
h≈ 0.693147
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Derivative of x 7→ 2x
Example
Let f (x) = 2x . Use a calculator to estimate f ′(0).
SolutionWe have
f ′(0) = limh→0
20+h − 20
h= lim
h→0
2h − 1
h≈ 0.693147
![Page 31: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/31.jpg)
Example
Use the previous fact to findd
dx2x .
Solution
d
dx2x = lim
h→0
2x+h − 2x
h= lim
h→0
2x2h − 2x
h
= limh→0
2x · 2h − 1
h
= 2x limh→0
2h − 1
h
≈ (0.693147)2x
Here we have a function whose derivative is a multiple of itself!(Much different from a polynomial.)
![Page 32: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/32.jpg)
Example
Use the previous fact to findd
dx2x .
Solution
d
dx2x = lim
h→0
2x+h − 2x
h= lim
h→0
2x2h − 2x
h
= limh→0
2x · 2h − 1
h
= 2x limh→0
2h − 1
h
≈ (0.693147)2x
Here we have a function whose derivative is a multiple of itself!(Much different from a polynomial.)
![Page 33: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/33.jpg)
Example
Use the previous fact to findd
dx2x .
Solution
d
dx2x = lim
h→0
2x+h − 2x
h= lim
h→0
2x2h − 2x
h
= limh→0
2x · 2h − 1
h
= 2x limh→0
2h − 1
h
≈ (0.693147)2x
Here we have a function whose derivative is a multiple of itself!(Much different from a polynomial.)
![Page 34: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/34.jpg)
Example
Findd
dx3x .
Solution
d
dx3x = lim
h→0
3x+h − 3x
h= lim
h→0
3x2h − 3x
h
= limh→0
3x · 3h − 1
h
= 3x limh→0
3h − 1
h
≈ (1.09861)3x
![Page 35: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/35.jpg)
Example
Findd
dx3x .
Solution
d
dx3x = lim
h→0
3x+h − 3x
h= lim
h→0
3x2h − 3x
h
= limh→0
3x · 3h − 1
h
= 3x limh→0
3h − 1
h
≈ (1.09861)3x
![Page 36: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/36.jpg)
TheoremLet a > 1, and let f (x) = ax . Then
f ′(x) = f ′(0)f (x)
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The natural exponential function
I If a = 2,d
dxax
∣∣∣∣x=0
< 1
I If a = 3,d
dxax
∣∣∣∣x=0
> 1
I We would hope there is a number a between 2 and 3 such
thatd
dxax
∣∣∣∣x=0
= 1
I We call this number e. Then by definition
d
dxex = ex
![Page 38: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/38.jpg)
Example
Findd
dx
(4x2 +
1
x+ 3 4√
x + 6ex
)
Solution
Remember1
x= x−1 and 4
√x = x1/4. So
dy
dx= 8x − 1
x2+
3
4x−3/4 + 6ex
![Page 39: Lesson 8: Derivatives of Polynomials and Exponential functions](https://reader033.vdocuments.mx/reader033/viewer/2022051313/5482f300b47959dd0c8b4976/html5/thumbnails/39.jpg)
Example
Findd
dx
(4x2 +
1
x+ 3 4√
x + 6ex
)
Solution
Remember1
x= x−1 and 4
√x = x1/4. So
dy
dx= 8x − 1
x2+
3
4x−3/4 + 6ex