Section 3.1Derivatives of Polynomials and Exponentials

Math 1a

February 20, 2008

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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

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