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)
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Image credit: Dot D
Outline
Recall
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of sine and cosine
Recall: the derivative
DefinitionLet f be a function and a a point in the domain of f . If the limit
f ′(a) = limh→0
f (a + h)− f (a)
h= lim
x→a
f (x)− f (a)
x − a
exists, the function is said to be differentiable at a and f ′(a) isthe derivative of f at a.The derivative . . .
I . . . measures the slope of the line through (a, f (a)) tangent tothe curve y = f (x);
I . . . represents the instantaneous rate of change of f at a
I . . . produces the best possible linear approximation to f near a.
Link between the notations
f ′(x) = lim∆x→0
f (x + ∆x)− f (x)
∆x= lim
∆x→0
∆y
∆x=
dy
dx
I Leibniz thought ofdy
dxas a quotient of “infinitesimals”
I We think ofdy
dxas representing a limit of (finite) difference
quotients
I The notation suggests things which are true even though theydon’t follow from the notation per se
Outline
Recall
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of sine and cosine
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.
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.
The squaring function and its derivatives
x
y
f
f ′f ′′ I f increasing =⇒ f ′ ≥ 0
I f decreasing =⇒ f ′ ≤ 0
I horizontal tangent at a=⇒ f ′(a) = 0
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) = 3x2.
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) = 3x2.
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 = 12 x−1/2.
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 = 12 x−1/2.
The square root function and its derivatives
x
y
f
f ′
I Here limx→0+
f ′(x) =∞and f is notdifferentiable at 0
I Notice alsolim
x→∞f ′(x) = 0
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) = 13 x−2/3.
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) = 13 x−2/3.
The cube root function and its derivatives
x
y
f
f ′
I Here limx→0
f ′(x) =∞ and
f is not differentiable at0
I Notice alsolim
x→±∞f ′(x) = 0
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)
= 13 x−2/3
(2x1/3
)= 2
3 x−1/3
So f ′(x) = 23 x−1/3.
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)
= 13 x−2/3
(2x1/3
)= 2
3 x−1/3
So f ′(x) = 23 x−1/3.
The function x 7→ x2/3 and its derivative
x
y
f
f ′
I f is not differentiable at0
I Notice alsolim
x→±∞f ′(x) = 0
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
Recap
y y ′
x2 2x
1
x3 3x2
x1/2 12 x−1/2
x1/3 13 x−2/3
x2/3 23 x−1/3
I The power goes down byone in each derivative
I The coefficient in thederivative is the power ofthe original function
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 .
Outline
Recall
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of sine and cosine
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.
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.
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.
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):
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