1.4 The Derivatives of Some Basic Functions

• The derivative function represents the slope of the tangent at each point on the graph of the function. (where it exists)

• Derivative of y=x.

x y Slope of tangent at x

-3 9 -6

-2 4 -4

-1 1 -2

0 0 0

1 1 2

2 4 4

3 9 6

2

2

2

2

2

2

•Slope of tangent at x =2x

•y’=2x

Derivative of y=x2

Derivative of y=x2

• The derivative of y=x2 is y’=2x

• See graph.

• See rate triangle

1

2=2x1

4=2x

The instantaneous rate of change of y with respect to x is 2x.

Visualize the point moving from left to right along the graph of y=x2 The slope along the tangent is always double the x-coordinate of the point.

y’=2x

Derivative of y=x3

x y=x3 Slope of tangent at x

-3 -27 27

-2 -8 12

-1 -1 3

0 0 0

1 1 3

2 8 12

3 27 27

-15

-9

-3

3

9

15

•Slope of tangent at x is 3x2

•y’=3x2

6

6

6

6

6

9

4

1

0

1

4

9

Derivative of y=x3

• The derivative of y=x3 is y’=3x2

• See graph.

• See rate triangle 13=3x2

The instantaneous rate of change of y with respect to x is 3x2.

Visualize the point moving from left to right along the graph of y=x3 . The slope along the tangent is always three time the square of the x-coordinate of the point.

Graph of y=x3 and its derivative

y ‘ = 3x2

Example 1• a) Find instantaneous rate of change of y=x2

at i) x=3 ii) x =-2

• b) Interpret graphically.

•Solution i)

•The derivative of y=x2 is y’=2x.

•When x=3, y’=2(3)=6

•y is increasing 6 times as fast as x is increasing when x=3.

m=6

Example 1• a) Find instantaneous rate of change of y=x2

at ii) x =-2

• b) Interpret graphically.

•Solution ii)

•The derivative of y=x2 is y’=2x.

•When x=-2, y’=2(-2)=-4

•y is decreasing 4 times as fast as x is increasing when x=-2.

m=-4

Example 2: Determine the equation of the tangent line.

• Determine the equation of the tangent to the graph of y=x3 at the point (-2,-8).

• Illustrate the results on a graph.•Solution

•The derivative of y=x3 is y’=3x2.

•When x=-2, the slope of the tangent line is y’=3(-2)2 = 12.

•The slope of the tangent is 12.

•The equation of the tangent line using y=m(x – p) + q is

•y=12(x+2)-8, or y=12x+16.

Example 2: Graph• Graph the function and the tangent line.

Derivative of y=mx+b

• y=x

• y=mx

• y=mx+b

• a) y’=1

• b) y’=m

• c) y’=m

Vertical Translation Rule• When the graph of a function is translated

vertically, the derivative is not affected.• The derivative of y=f(x)+c is y’=f ’(x),

where c is any constant.• Example: Find the derivative of y=x2+4.• y’=2x.

Vertical Translation RuleExample: Find the derivative of y=x2+4.• y’=2x.

Vertical Stretch Rule

• When the graph of a function is expanded or compressed vertically, the graph of the derivative is also expanded or compressed vertically.

• The derivative of y=cf(x) is y’=cf’(x), where c is any constant.

Vertical Stretch Rule• Example: Find the derivative of y=2x3

• f(x)=x3 , f’(x)=3x2

• y’=2f’(x) , so y’=2(3x2)• Or y’=6x2

•When the function is stretched by a factor of 2, the y-values for any x-value are doubled.

•That means when the slope is calculated at any point the slope will be doubled.

Practice• Find the derivatives of the following

functions.

• a) y=x3+ 23

• b) y=-5x2

• c) y=4x3-7

• d) y=7x

• e) y=14 - x3

f(x)=x2 ;f ‘(x)=2x f(x)=x3 ;f ‘(x)=3x2

•a) y’=3x2

•b) y’=-10x

•c) y’=12x2

•d) y’=7

•e) y’=-3x2

If y = f(x) +c, then y ‘= f ‘(x) If y = cf(x), then y ‘= cf ‘(x)