Differentiability

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Let’s clear up some confusion…

To find the derivative of f(x)

EX: Find if

To find the derivative at a point a of f(x)

EX: Find if

𝐥𝐢𝐦𝒉→𝟎

𝒇 (𝒙+𝒉) − 𝒇 (𝒙 )𝒉

𝐥𝐢𝐦𝒙→𝒂

𝒇 (𝒙 ) − 𝒇 (𝒂)𝒙−𝒂

𝐥𝐢𝐦𝒙→𝒂

𝒇 (𝒙 ) − 𝒇 (𝒂)𝒙−𝒂

=𝐥𝐢𝐦𝒙→𝟓

𝒙𝟐−𝟐𝟓𝒙−𝟓

=𝐥𝐢𝐦𝒙→𝟓

(𝒙+𝟓)(𝒙−𝟓)(𝒙−𝟓)

=𝐥𝐢𝐦𝒙 →𝟓

(𝒙+𝟓)=𝟏𝟎

In General!

Specific!

To be differentiable, a function must be continuous and smooth.

Derivatives will fail to exist at:

corner cusp

vertical tangent discontinuity

f x x 2

3f x x

3f x x 1, 0

1, 0

xf x

x

Two theorems:

If f has a derivative at x = a, then f is continuous at x = a.

Since a function must be continuous to have a derivative, if it has a derivative then it is continuous.

**The converse of this theorem is false!

1

2f a

3f b

Intermediate Value Theorem for Derivatives

Between a and b, must take

on every value between and .

f 1

23

If a and b are any two points in an interval on which f is

differentiable, then takes on every value between

and .

f f a

f b

p

If the derivative of a function is its slope, then for a constant function, the derivative must be zero.

0d

cdx

example: 3y

0y

The derivative of a constant is zero.

Rules for Differentiation Introduction

If we find derivatives with the difference quotient:

2 22

0limh

x h xdx

dx h

2 2 2

0

2limh

x xh h x

h

2x

3 33

0limh

x h xdx

dx h

3 2 2 3 3

0

3 3limh

x x h xh h x

h

23x

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(Pascal’s Triangle)

4dx

dx

4 3 2 2 3 4 4

0

4 6 4limh

x x h x h xh h x

h

34x

We observe a pattern: 2x 23x 34x 45x 56x …

1n ndx nx

dx

examples:

4f x x

34f x x

8y x

78y x

power rule

We observe a pattern: 2x 23x 34x 45x 56x …

d ducu c

dx dx

examples:

1n ndcx cnx

dx

constant multiple rule:

5 4 47 7 5 35d

x x xdx

When we used the difference quotient, we observed that since the limit had no effect on a constant coefficient, that the constant could be factored to the outside.

(Each term is treated separately)

d ducu c

dx dx

constant multiple rule:

sum and difference rules:

d du dvu v

dx dx dx d du dv

u vdx dx dx

4 12y x x 34 12y x

4 22 2y x x

34 4dy

x xdx

This makes sense, because:

0limh

f x h g x h f x g x

h

0 0

lim limh h

f x h f x g x h g x

h h

Example:Find the horizontal tangents of: 4 22 2y x x

34 4dy

x xdx

Horizontal tangents occur when slope = zero.34 4 0x x

3 0x x

2 1 0x x

1 1 0x x x

0, 1, 1x

Plugging the x values into the original equation, we get:

2, 1, 1y y y

(The function is even, so we only get two horizontal tangents.)

4 22 2y x x

4 22 2y x x

2y

4 22 2y x x

2y

1y

4 22 2y x x

4 22 2y x x

First derivative (slope) is zero at:

0, 1, 1x

34 4dy

x xdx