8.11: Quotient Rule and Higher Derivatives
You may recall hearing about Becky and her Track and Field competition in a prior lesson. Her boyfriend had taken a picture of her just as she started to pull away from the others on the track. We learned how she might learn to identify her instantaneous speed at just the split second the picture was taken by using Calculus to find a derivative.
What if, instead of just finding her speed at that split second, she wanted to find her acceleration ?
Watch This
Embedded Video:
- James Sousa: Higher-Order Derivatives
Guidance
The Quotient Rule
Theorem: (The Quotient Rule)
If
f
and
g
are differentiable functions at
x
and
g
(
x
) ≠ 0, then
In simpler notation Keep in mind that the order of operations is important (because of the minus sign in the numerator) and . |
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Higher Derivatives
If the derivative of the function is differentiable, then the derivative of , denoted by is called the second derivative of . We can continue the process of differentiating derivatives and obtain third, fourth, fifth and higher derivatives of . They are denoted by , , , , , . . . ,
Example A
Find for
Solution
Example B
At which point(s) does the graph of have a horizontal tangent line?
Solution
Since the slope of a horizontal line is zero, and since the derivative of a function signifies the slope of the tangent line, then taking the derivative and equating it to zero will enable us to find the points at which the slope of the tangent line is equal to zero, i.e., the locations of the horizontal tangents. Notice that we will need to use the quotient rule here:
- Multiply both sides by ,
- Therefore, at and , the tangent line is horizontal.
Example C
Find the fifth derivative of
Solution
To find the fifth derivative, we must first find the first, second, third, and fourth derivatives.
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Concept question wrap-up Once Becky has calculated her instantaneous speed at a given point on the track by finding the derivative, she could then take the derivative of that function to find her instantaneous acceleration at the same point in the race. By finding her instantaneous speed and acceleration at different points in the race, she can learn a lot about what points made a difference in her overall success, and also what points she needs to work on. Better racing through Calculus! |
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Vocabulary
Instantaneous velocity is speed in a given direction at a single "snapshot" moment in time.
Instantaneous acceleration is change in velocity calculated at a single instant.
Guided Practice
Questions
1) Suppose y'(2) = 0 and (y/q)(2) = 0. Find q(2) assuming y(2) = 0
2) Find the derivative of
3) Given What is ?
4) Given Find when
Solutions
1) Begin with the quotient rule:
- ..... Substitute
- ..... Substituting again with given values
- ..... Simplify with:
2) Use the quotient rule: Note: and
- ..... Substitute
- ..... Simplify
3) Recall that means "The derivative of the derivative of x "
- ..... By the power rule
4) Recall that means "The derivative of the derivative of x "
- ..... Use the power rule on f(x)
- ..... Use the power rule on f'(x)
- ..... Substitute 3
Practice
Use the Quotient Rule to Solve:
- Suppose and . Find assuming
- Given: What is: ?
- Given: what is ?
- What is ?
Solve these Higher Order Derivatives:
- Given: What is ?
- Given: What is ?
- Given: What is ?
- Given: What is ?
- What is ?
Solve:
- Newton’s Law of Universal Gravitation states that the gravitational force between two masses (say, the earth and the moon), m and M is equal two their product divided by the squared of the distance r between them. Mathematically, where G is the Universal Gravitational Constant (1.602 × 10 ^{ -11 } Nm ^{ 2 } /kg ^{ 2 } ). If the distance r between the two masses is changing, find a formula for the instantaneous rate of change of F with respect to the separation distance r .
- Find : where is a constant.
- Find where .
Image Attributions
Description
Learning Objectives
Here you will learn how to find the derivative of a function which is divided by another function. You will also explore taking the derivatives of derivatives.
Difficulty Level:
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Date Created:
Nov 01, 2012Last Modified:
Nov 24, 2014Vocabulary
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