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Derivative as a function Domain of f ′(x) Example Graph of the derivative f ′(x) Domain of Derivative of |x| Different Notation Differentiability Differentiable implies continuous Points where function is not differentiable Higher Derivatives Acceleration Notation for Higher derivatives

Derivative as a function

In the previous section we defined the derivative of a function f at a number a(when the function f is defined in an open interval containing a) to be

f ′(a) = limh→0

f (a + h)− f (a)

when this limit exists. This gives the slope of the tangent to the curvey = f (x) when x = a

Example Last day we saw that if f (x) = x2 + 5x , then f ′(a) = 2a + 5 for anyvalue of a. Therefore f ′(1) = 7, f ′(2) = 9, f ′(2.5) = 10 etc....

The value of f ′(a) varies as the number a varies, hence f ′ is a function of a.We can change the variable from a to x to get a new function, called Thederivative of f

f ′(x) = limh→0

f (x + h)− f (x)

Annette Pilkington Lecture 23 : Sequences

Domain of f ′(x)

f ′(x) = limh→0

f (x + h)− f (x)

(f ′(x) is defined when f is defined in an open interval containing x and theabove limit exists). Note that when calculating this limit for a particular valueof x , h→ 0 and the value of x remains constant.

Note also that if x is in the domain of f ′, it must satisfy the following 3conditions:

1. x must be in the domain of f .

2. limh→0f (x+h)−f (x)

hmust exist at x .

3. f must be defined in an open interval containing x .

The domain of the function f ′ may be smaller than the domain of the function

f since 2 or 3 may fail for some values of x in the domain of f .

Example

Example What is f ′(x) when f (x) = x2 + 2x + 4?. What is the domain off ′(x)?

I To calculate f ′(x), we calculate using limits as we did for f ′(a) in theprevious lecture, replacing a by x .

I f ′(x) = limh→0f (x+h)−f (x)

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

I = limh→0h(2x+h+2)

h= limh→0 �h(2x+h+2)

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

I Since the domain of f is all real numbers and the above limit exists for allreal numbers, the domain of f ′ is also all real numbers.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Example

I = limh→0(x+h)2+2(x+h)+4−[x2+2x+4]

I = limh→0x2+2xh+h2+2x+2h+4−x2−2x−4

I = limh→0�x2+2xh+h2+�2x+2h+�4−�x

2−�2x−�4h

= limh→02xh+h2+2h

�h= limh→0(2x + h + 2) = 2x + 2.

I f ′(x) = 2x + 2.

Graph of the derivative f ′(x)

Below we see how the graph of f (x) = x2 + 2x + 4 is related to the graph of itsderivative f ′(x) = 2x + 2, which gives the slope of the tangents to the graph off (x) = x2 + 2x + 4. (See Mathematica File)

Fill in <, > or = as appropriate:

When f (x) is decreasing the function f ′(x) 0

When f (x) is increasing the function f ′(x) 0

At the turning point x = −1, f ′(x) 0

I When f (x) is decreasing the function f ′(x) < 0

I When f (x) is increasing the function f ′(x) > 0

I At the turning point x = −1, f ′(x) = 0

Domain of Derivative of |x |Consider the function f (x) = |x |.Does limh→0

f (x+h)−f (x)h

exist when x > 0?

I limh→0f (x+h)−f (x)

h= limh→0

|x+h|−|x|h

I When x > 0, both x + h and x are > 0, when h is very close to 0.

I So limh→0|x+h|−|x|

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

Does limh→0f (x+h)−f (x)

hexist when x < 0?

h= limh→0

|x+h|−|x|h

I When x < 0, both x + h and x are < 0, when h is very close to 0.

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

does not exist since the limit from the left is −1 and thelimit from the right is +1. (no tangent at 0 on graph)

What is the domain of f ′(x)?

I Domain f ′(x) = all real numbers except 0.

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

f (x+h)−f (x)h

exist when x > 0?

h= limh→0

|x+h|−|x|h

h= limh→0

x+h−xh

= limh→0hh

= limh→0 1 = 1.

hexist when x < 0?

h= limh→0

|x+h|−|x|h

h= limh→0

−(x+h)+xh

= limh→0−hh

= limh→0(−1) = −1.

hexist when x = 0?

I limh→0f (0+h)−f (0)

h= limh→0

|h|−|0|h

= limh→0|h|h

I |h|h

= 1 if h > 0 and |h|h

= −1 if h < 0.

I So limh→0|h|h

Different Notation

Using y = f (x), to denote that the independent variable is y , there are anumber of notations used to denote the derivative of f (x) :

f ′(x) = y ′ =dy

dxf (x) = Df (x) = Dx f (x).

The symbols D and ddx

are called differential operators, because when they areapplied to a function, they transform the function to its derivative. The symboldydx

should not be interpreted as a quotient rather it is a limit originating fromthe notation

dx= lim

∆x→0

When we evaluate the derivative at a number a, we use the following notation

f ′(a) =dy

˛̨̨̨˛x=a

Differentiability

Definition When a function f is defined in an open interval containing a, wesay a function f is differentiable at a if f ′(a) exists. [ That is, conditions 1, 2and 3 from page 1 must be satisfied for x = a.] It is differentiable on an openinterval , (a, b) (or (a,∞) or (−∞, a)) if it is differentiable at every number inthe interval.

Note: Saying that f is differentiable at a is the same as saying that a is in thedomain of f ′.Example Let f (x) = |x |. Is f (x) differentiable at 0?

I No because, as we saw above, f ′(0) does not exist.

If f (x) differentiable on the intervals (−∞, 0) and (0,∞).

I yes because, f ′(x) exists for all values of x in the intervals (−∞, 0) and(0,∞).

Is f (x) continuous at 0?

I yes f (x) = |x | is continuous at 0, because limx→0 |x | = 0. However, as weshowed above, it is not differentiable at 0. (geometrically: there is a sharppoint on the curve and no tangent line exists).

Differentiability

Differentiable at a implies continuous at a

Theorem If f is differentiable at a, then f is continuous at a.

In particular the theorem shows that if a function has a discontinuity at a pointa, then it cannot be differentiable at a. (Note by the previous example, theconverse is not true; a function can be continuous at a, but not differentiableat a).

Geometrically, a function is differentiable at a point a if its graph is smooth ata. A function f can fail to be differentiable at a point a in a number of ways.

I The function might be continuous at a, but have a sharp point or kink inthe graph, like the graph of f (x) = |x | at 0.

I The function might not be continuous or might be undefined at a.

I The function might be continuous but the tangent line may be vertical,i.e. limx→a

f (x)−f (a)x−a

= ±∞.

f (x)−f (a)x−a

= ±∞.

f (x)−f (a)x−a

= ±∞.

f (x)−f (a)x−a

= ±∞.

Points where function is not differentiable

Example Identify the points in the graphs below where the functions are notdifferentiable.

(1.1)(1.1)

(1.2)(1.2)

Exampleswith(plots,implicitplot);

implicitplot

Plot a circle.

T:=implicitplot(x^2=1, x=-1..1, y=-1..1);T := PLOT ...

implicitplot y3

= x2, x =K10 ..10, y =K5 ..5, gridrefine = 5 ;

K10 K5 0 5 10

– 10

– 12

– 14

– 16

15 – 10 – 5 5 10 15

f x( ) = x

I The graph on the left has a sharp point at x = 0. So the function is notdifferentiable at x = 0.

I The graph in the center has a vertical tangent at x = 0. So the functionis not differentiable at x = 0.

I The graph on the right is not continuous at x = 3, x = 5, x = 7 andx = 10. So the function cannot be differentiable at those points.

(1.1)(1.1)

(1.2)(1.2)

implicitplot

Plot a circle.

implicitplot y3

K10 K5 0 5 10

– 10

– 12

– 14

– 16

15 – 10 – 5 5 10 15

f x( ) = x

(1.1)(1.1)

(1.2)(1.2)

implicitplot

Plot a circle.

implicitplot y3

K10 K5 0 5 10

– 10

– 12

– 14

– 16

15 – 10 – 5 5 10 15

f x( ) = x

(1.1)(1.1)

(1.2)(1.2)

implicitplot

Plot a circle.

implicitplot y3

K10 K5 0 5 10

– 10

– 12

– 14

– 16

15 – 10 – 5 5 10 15

f x( ) = x

Higher Derivatives

We have seen that given a function f (x), we can define a new function f ′(x).We can continue this process by defining a new function,

f ′′(x) =d

dxf ′(x).

This is the second derivative of the function f (x). This function gives the slopeof the tangent to the curve y = f ′(x) at each value of x .We can then define the third derivative of f (x) as the derivative of the secondderivative, etc...

Example Let f (x) = x2 + 2x + 4. We saw above that the derivative of f (x)is f ′(x) = 2x + 2. Find and interpret the second derivative of f (x);

I f ′′(x) = limh→0f ′(x+h)−f ′(x)

h= limh→0

2(x+h)+2−(2x+2)h

I = limh→02x+2h+2−2x−2

h= limh→0 �2x+2h+�2−�2x−�2

I = limh→02hh

= limh→0 2 = 2.

I f ′′(x) = 2 for all values of x . (which makes sense, since f ′′(x) is the slopeof the tangent to the graph of f ′(x) = 2x + 2 for any x).

Higher Derivatives

f ′′(x) =d

dxf ′(x).

I f ′′(x) = limh→0f ′(x+h)−f ′(x)

h= limh→0

2(x+h)+2−(2x+2)h

I = limh→02x+2h+2−2x−2

h= limh→0 �2x+2h+�2−�2x−�2

I = limh→02hh

= limh→0 2 = 2.

Higher Derivatives

f ′′(x) =d

dxf ′(x).

I f ′′(x) = limh→0f ′(x+h)−f ′(x)

h= limh→0

2(x+h)+2−(2x+2)h

I = limh→02x+2h+2−2x−2

h= limh→0 �2x+2h+�2−�2x−�2

I = limh→02hh

= limh→0 2 = 2.

Higher Derivatives

f ′′(x) =d

dxf ′(x).

I f ′′(x) = limh→0f ′(x+h)−f ′(x)

h= limh→0

2(x+h)+2−(2x+2)h

I = limh→02x+2h+2−2x−2

h= limh→0 �2x+2h+�2−�2x−�2

I = limh→02hh

= limh→0 2 = 2.

Higher Derivatives

f ′′(x) =d

dxf ′(x).

I f ′′(x) = limh→0f ′(x+h)−f ′(x)

h= limh→0

2(x+h)+2−(2x+2)h

I = limh→02x+2h+2−2x−2

h= limh→0 �2x+2h+�2−�2x−�2

I = limh→02hh

= limh→0 2 = 2.

Acceleration

The second derivative gives us the rate of change of the rate of change. In thecase of a position function s = s(t) of an object moving in a straight line, thederivative v(t) = s ′(t) gives us the velocity of the moving object at time t andthe second derivative a(t) = v ′(t) = s ′′(t) gives us the acceleration of themoving object at time t. This is the rate of change of the velocity at time t.

Example The position of an object moving in a straight line at time t is givenby s(t) = t2 + 2t + 4. What is the velocity and acceleration of the object aftert = 5 seconds?

I If s(t) = t2 + 2t + 4, then the velocity is given by its derivatives ′(t) = v(t) = 2t + 2.we’ve already worked through the calculation with the variable x insteadof t.

I The velocity after 5 seconds is v(5) = 2(5) + 2 = 12 m/s (if distance ismeasured in meters).

I The acceleration function is the derivative of the velocity function.a(t) = v ′(t) = 2. (We worked through this calculation with the variable xabove.)

I The acceleration is constant and a(5) = 2.

Acceleration

Notation for Higher derivatives

The second derivative is also denoted by

f ′′(x) =d

dx2= y ′′.

The third derivative of f is the derivative of the second derivative, denoted

dxf ′′(x) = f ′′′(x) = y ′′′ = y (3) =

“d2y

Higher derivative are denoted

f (4)(x) = y (4) =d4y

dx4, f (5)(x) = y (5) =

dx5, etc . . .

Example If f (x) = x2 + 2x + 4, find f (4)(x) and f (5)(x).

I Above, we saw that f ′′(x) = 2 for all x .

I f ′′′(x) = df ′′(x)dx

= limh→0f ′′(x+h)−f ′′(x)

h= limh→0

2−2h

= limx→00h

I f (4)(x) = df ′′′(x)dx

= limh→0f ′′′(x+h)−f ′′′(x)

h= limh→0

0−0h

= limx→00h

I f (5)(x) = df (4)(x)dx

= limh→0f (4)(x+h)−f (4)(x)

h= limh→0

0−0h

= limx→00h

f ′′(x) =d

dx2= y ′′.

dxf ′′(x) = f ′′′(x) = y ′′′ = y (3) =

“d2y

f (4)(x) = y (4) =d4y

dx4, f (5)(x) = y (5) =

dx5, etc . . .

I f ′′′(x) = df ′′(x)dx

= limh→0f ′′(x+h)−f ′′(x)

h= limh→0

2−2h

= limx→00h

I f (4)(x) = df ′′′(x)dx

= limh→0f ′′′(x+h)−f ′′′(x)

h= limh→0

0−0h

= limx→00h

I f (5)(x) = df (4)(x)dx

= limh→0f (4)(x+h)−f (4)(x)

h= limh→0

0−0h

= limx→00h

f ′′(x) =d

dx2= y ′′.

dxf ′′(x) = f ′′′(x) = y ′′′ = y (3) =

“d2y

f (4)(x) = y (4) =d4y

dx4, f (5)(x) = y (5) =

dx5, etc . . .

I f ′′′(x) = df ′′(x)dx

= limh→0f ′′(x+h)−f ′′(x)

h= limh→0

2−2h

= limx→00h

I f (4)(x) = df ′′′(x)dx

= limh→0f ′′′(x+h)−f ′′′(x)

h= limh→0

0−0h

= limx→00h

I f (5)(x) = df (4)(x)dx

= limh→0f (4)(x+h)−f (4)(x)

h= limh→0

0−0h

= limx→00h

f ′′(x) =d

dx2= y ′′.

dxf ′′(x) = f ′′′(x) = y ′′′ = y (3) =

“d2y

f (4)(x) = y (4) =d4y

dx4, f (5)(x) = y (5) =

dx5, etc . . .

I f ′′′(x) = df ′′(x)dx

= limh→0f ′′(x+h)−f ′′(x)

h= limh→0

2−2h

= limx→00h

I f (4)(x) = df ′′′(x)dx

= limh→0f ′′′(x+h)−f ′′′(x)

h= limh→0

0−0h

= limx→00h

I f (5)(x) = df (4)(x)dx

= limh→0f (4)(x+h)−f (4)(x)

h= limh→0

0−0h

= limx→00h

f ′′(x) =d

dx2= y ′′.

dxf ′′(x) = f ′′′(x) = y ′′′ = y (3) =

“d2y

f (4)(x) = y (4) =d4y

dx4, f (5)(x) = y (5) =

dx5, etc . . .

I f ′′′(x) = df ′′(x)dx

= limh→0f ′′(x+h)−f ′′(x)

h= limh→0

2−2h

= limx→00h

I f (4)(x) = df ′′′(x)dx

= limh→0f ′′′(x+h)−f ′′′(x)

h= limh→0

0−0h

= limx→00h

I f (5)(x) = df (4)(x)dx

= limh→0f (4)(x+h)−f (4)(x)

h= limh→0

0−0h

= limx→00h

f0 x Domain of Derivative of Derivative as a functionapilking/Math10550/Lectures/Lecture7.pdf(f0(x)...

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