Section 2.3 Basic Differentiation Rules V63.0121, Calculus I February 11–12, 2009 Announcements I new OH: M 1–2 (Calc only), T 1–2, W 2–3 (calc only, after 2/11), R 9–10am I Quiz next week on Sections 1.3–1.6 I Midterm March 4 or 5 (75 min., in class, Sections 1.1.–2.4) I ALEKS is due February 27 at 11:59pm
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Section 2.3Basic Differentiation Rules
V63.0121, Calculus I
February 11–12, 2009
Announcements
I new OH: M 1–2 (Calc only), T 1–2, W 2–3 (calc only, after2/11), R 9–10am
I Quiz next week on Sections 1.3–1.6
I Midterm March 4 or 5 (75 min., in class, Sections 1.1.–2.4)
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.
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)
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.
Pascal’s Triangle
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’s Triangle
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’s Triangle
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’s Triangle
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 (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.
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.
The Power Rule for constants
TheoremLet c be a constant. Then
d
dxc = 0
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.
The Power Rule for constants
TheoremLet c be a constant. Then
d
dxc = 0
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.
The Power Rule for constants
TheoremLet c be a constant. Then
d
dxc = 0
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.
New derivatives from oldThis is where the calculus starts to get really powerful!
Calculus
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 ′.
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.
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) = c · f ′(x)
Succinctly, (cf )′ = cf ′.
Proof.Again, follow your nose.
(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)
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
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
Outline
Recall
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of sine and cosine
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x .
Proof.From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
(sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x .
Proof.From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
(sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
Angle addition formulasSee Appendix A
sin(A + B) = sin A cos B + cos A sin B
cos(A + B) = cos A cos B − sin A sin B
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x .
Proof.From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
(sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x .
Proof.From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
(sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
Two important trigonometric limitsSee Section 1.4
θ
sin θ
1− cos θ
θ
−1 1
limθ→0
sin θ
θ= 1
limθ→0
cos θ − 1
θ= 0
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x .
Proof.From the definition:
d
dxsin x = lim
h→0
sin(x + h)− sin x
h
= limh→0
(sin x cos h + cos x sin h)− sin x
h
= sin x · limh→0
cos h − 1
h+ cos x · lim
h→0
sin h
h
= sin x · 0 + cos x · 1 = cos x
Illustration of Sine and Cosine
x
y
π −π2
0 π2
π
sin x
cos x
I f (x) = sin x has horizontal tangents where f ′ = cos(x) is zero.
I what happens at the horizontal tangents of cos?
Illustration of Sine and Cosine
x
y
π −π2
0 π2
π
sin xcos x
I f (x) = sin x has horizontal tangents where f ′ = cos(x) is zero.
I what happens at the horizontal tangents of cos?
Illustration of Sine and Cosine
x
y
π −π2
0 π2
π
sin xcos x
I f (x) = sin x has horizontal tangents where f ′ = cos(x) is zero.
I what happens at the horizontal tangents of cos?
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.We already did the first. The second is similar (mutatis mutandis):
d
dxcos x = lim
h→0
cos(x + h)− cos x
h
= limh→0
(cos x cos h − sin x sin h)− cos x
h
= cos x · limh→0
cos h − 1
h− sin x · lim
h→0
sin h
h
= cos x · 0− sin x · 1 = − sin x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.We already did the first. The second is similar (mutatis mutandis):
d
dxcos x = lim
h→0
cos(x + h)− cos x
h
= limh→0
(cos x cos h − sin x sin h)− cos x
h
= cos x · limh→0
cos h − 1
h− sin x · lim
h→0
sin h
h
= cos x · 0− sin x · 1 = − sin x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.We already did the first. The second is similar (mutatis mutandis):
d
dxcos x = lim
h→0
cos(x + h)− cos x
h
= limh→0
(cos x cos h − sin x sin h)− cos x
h
= cos x · limh→0
cos h − 1
h− sin x · lim
h→0
sin h
h
= cos x · 0− sin x · 1 = − sin x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.We already did the first. The second is similar (mutatis mutandis):
d
dxcos x = lim
h→0
cos(x + h)− cos x
h
= limh→0
(cos x cos h − sin x sin h)− cos x
h
= cos x · limh→0
cos h − 1
h− sin x · lim
h→0
sin h
h
= cos x · 0− sin x · 1 = − sin x
Derivatives of Sine and Cosine
Fact
d
dxsin x = cos x
d
dxcos x = − sin x
Proof.We already did the first. The second is similar (mutatis mutandis):