lesson 8: derivatives of polynomials and exponential functions
DESCRIPTION
Some of the most famous rules of the calculus of derivatives: the power rule, the sum rule, the constant multiple rule, and the number e defined so that e^x is its own derivative!TRANSCRIPT
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)
Outline
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
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.
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.
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.
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.
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.
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.
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.
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.
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.
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
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
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.
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
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.
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.
New derivatives from old
This is where the calculus starts to get really powerful!
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) = cf ′(x)
Succinctly, (cf )′ = cf ′.
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)
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
Derivatives so far
Derivatives of polynomialsThe power rule for whole numbersLinear combinations
Derivatives of exponential functionsBy experimentationThe natural exponential functionFinal examples
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
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
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.)
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.)
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.)
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
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
TheoremLet a > 1, and let f (x) = ax . Then
f ′(x) = f ′(0)f (x)
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
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
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