lesson 9: basic differentiation rules

Section 2.3Basic Differentiation Rules

V63.0121, Calculus I

February 11–12, 2009

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Outline

Recall

Derivatives so far

Derivatives of polynomialsThe power rule for whole numbersLinear combinations

Derivatives of sine and cosine

Derivative

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

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.

Notation

Newtonian notation Leibnizian notation

f ′(x) y ′(x) y ′dy

dxf (x)

Link between the notations

f ′(x) = lim∆x→0

f (x + ∆x)− f (x)

∆x= lim

∆x→0

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

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

(x + h)2 − x2

= limh→0

��x2 + 2xh + h2 −��x

h= lim

2x�h + h�2

= limh→0

(2x + h) = 2x .

So f ′(x) = 2x.

Derivative of the squaring function

Example

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

(x + h)2 − x2

= limh→0

��x2 + 2xh + h2 −��x

h= lim

2x�h + h�2

= limh→0

(2x + h) = 2x .

So f ′(x) = 2x.

The squaring function and its derivatives

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

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

(x + h)3 − x3

= limh→0

��x3 + 3x2h + 3xh2 + h3 −��x

h= lim

3x2�h + 3xh��

2 + h��2

= limh→0

(3x2 + 3xh + h2

)= 3x2.

So f ′(x) = 3x2.

Derivative of the cubing function

Example

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

(x + h)3 − x3

= limh→0

��x3 + 3x2h + 3xh2 + h3 −��x

h= lim

3x2�h + 3xh��

2 + h��2

= limh→0

(3x2 + 3xh + h2

)= 3x2.

So f ′(x) = 3x2.

The cubing function and its derivatives

f ′f ′′

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

√x + h −

= limh→0

√x + h −

h·√

x + h +√

x√x + h +

= limh→0

(�x + h)−�x

x + h +√

x) = lim

�h(√

x + h +√

So f ′(x) =√

x = 12 x−1/2.

Derivative of the square root function

Example

Suppose f (x) =√

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

√x + h −

= limh→0

√x + h −

h·√

x + h +√

x√x + h +

= limh→0

(�x + h)−�x

x + h +√

x) = lim

�h(√

x + h +√

So f ′(x) =√

x = 12 x−1/2.

The square root function and its derivatives

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√

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

(x + h)1/3 − x1/3

= 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((x + h)2/3 + (x + h)1/3x1/3 + x2/3

So f ′(x) = 13 x−2/3.

Derivative of the cube root function

Example

Suppose f (x) = 3√

Solution

f ′(x) = limh→0

f (x + h)− f (x)

h= lim

(x + h)1/3 − x1/3

= 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((x + h)2/3 + (x + h)1/3x1/3 + x2/3

So f ′(x) = 13 x−2/3.

The cube root function and its derivatives

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

(x + h)2/3 − x2/3

= limh→0

(x + h)1/3 − x1/3

(x + h)1/3 + x1/3)

= 13 x−2/3

(2x1/3

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

(x + h)2/3 − x2/3

= limh→0

(x + h)1/3 − x1/3

(x + h)1/3 + x1/3)

= 13 x−2/3

(2x1/3

3 x−1/3

So f ′(x) = 23 x−1/3.

The function x 7→ x2/3 and its derivative

I f is not differentiable at0

I Notice alsolim

x→±∞f ′(x) = 0

y y ′

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

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x2 2x1

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x2 2x1

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−1/3

y y ′

x3 3x2

x1/2 12 x−1/2

x1/3 13 x−2/3

x2/3 23 x−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

Recall

Derivatives so far

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

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 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 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 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 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

. . . . . .

Let r be a positive whole number. Then

dxx r = rx r−1

Proof.As we showed above,

(x + h)n − xn

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.

Let r be a positive whole number. Then

dxx r = rx r−1

Proof.As we showed above,

(x + h)n − xn

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

dxc = 0

dxx0 = 0x−1

(although x 7→ 0x−1 is not defined at zero.)

Proof.Let f (x) = c . Then

f (x + h)− f (x)

c − c

So f ′(x) = limh→0

0 = 0.

dxc = 0

dxx0 = 0x−1

f (x + h)− f (x)

c − c

0 = 0.

dxc = 0

dxx0 = 0x−1

f (x + h)− f (x)

c − c

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)

= limh→0

f (x + h) + g(x + h)− [f (x) + g(x)]

= limh→0

f (x + h)− f (x)

h+ lim

g(x + h)− g(x)

= 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)

= limh→0

cf (x + h)− cf (x)

= c limh→0

f (x + h)− f (x)

= c · f ′(x)

Derivatives of polynomials

Example

(2x3 + x4 − 17x12 + 37

Solution

(2x3 + x4 − 17x12 + 37

dxx4 +

(−17x12

dx(37)

dxx3 +

dxx4 − 17

dxx12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

Derivatives of polynomials

Example

(2x3 + x4 − 17x12 + 37

)Solution

(2x3 + x4 − 17x12 + 37

dxx4 +

(−17x12

dx(37)

dxx3 +

dxx4 − 17

dxx12 + 0

= 2 · 3x2 + 4x3 − 17 · 12x11

= 6x2 + 4x3 − 204x11

Outline

Recall

Derivatives so far

Derivatives of Sine and Cosine

dxsin x = cos x .

Proof.From the definition:

dxsin x = lim

sin(x + h)− sin x

= limh→0

(sin x cos h + cos x sin h)− sin x

= sin x · limh→0

cos h − 1

h+ cos x · lim

= sin x · 0 + cos x · 1 = cos x

dxsin x = cos x .

dxsin x = lim

sin(x + h)− sin x

= limh→0

= sin x · limh→0

cos h − 1

h+ cos x · lim

= 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

dxsin x = cos x .

dxsin x = lim

sin(x + h)− sin x

= limh→0

= sin x · limh→0

cos h − 1

h+ cos x · lim

= sin x · 0 + cos x · 1 = cos x

dxsin x = cos x .

dxsin x = lim

sin(x + h)− sin x

= limh→0

= sin x · limh→0

cos h − 1

h+ cos x · lim

= sin x · 0 + cos x · 1 = cos x

Two important trigonometric limitsSee Section 1.4

sin θ

1− cos θ

−1 1

limθ→0

sin θ

limθ→0

cos θ − 1

dxsin x = cos x .

dxsin x = lim

sin(x + h)− sin x

= limh→0

= sin x · limh→0

cos h − 1

h+ cos x · lim

= sin x · 0 + cos x · 1 = cos x

Illustration of Sine and Cosine

π −π2

I f (x) = sin x has horizontal tangents where f ′ = cos(x) is zero.

I what happens at the horizontal tangents of cos?

π −π2

sin xcos x

π −π2

sin xcos x

dxsin x = cos x

dxcos x = − sin x

Proof.We already did the first. The second is similar (mutatis mutandis):

dxcos x = lim

cos(x + h)− cos x

= limh→0

(cos x cos h − sin x sin h)− cos x

= cos x · limh→0

cos h − 1

h− sin x · lim

= cos x · 0− sin x · 1 = − sin x

dxsin x = cos x

dxcos x = − sin x

dxcos x = lim

cos(x + h)− cos x

= limh→0

= cos x · limh→0

cos h − 1

h− sin x · lim

= cos x · 0− sin x · 1 = − sin x

dxsin x = cos x

dxcos x = − sin x

dxcos x = lim

cos(x + h)− cos x

= limh→0

= cos x · limh→0

cos h − 1

h− sin x · lim

= cos x · 0− sin x · 1 = − sin x

dxsin x = cos x

dxcos x = − sin x

dxcos x = lim

cos(x + h)− cos x

= limh→0

= cos x · limh→0

cos h − 1

h− sin x · lim

= cos x · 0− sin x · 1 = − sin x

dxsin x = cos x

dxcos x = − sin x

dxcos x = lim

cos(x + h)− cos x

= limh→0

= cos x · limh→0

cos h − 1

h− sin x · lim

= cos x · 0− sin x · 1 = − sin x

lesson 9: basic differentiation rules

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