Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
a) What is the gradient of the line \(y=-2x+5\)?
b) What is the gradient of the line segment connecting \(|\left(2,5\right)\) and \(B\left(7,0\right)\)?
Solutions
Get Ready
Questions
What do we mean by the gradient of a curve?
Solutions
Notes
The derivative or gradient function is the function whose image at \(x=c\) is the gradient of \(f\left(x\right)\) at \(c\). The notation for the gradient function is \(f’\left(x\right)\) which is read as ‘\(f\) dashed of \(x\)’ (UK) or ‘\(f\) prime of \(x\)’ (USA).
Examples and Your Turns
Example
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Your Turn
Sketch the derivative of the following function.
Always / Sometimes / Never
| # | Statement | Answer |
|---|---|---|
| 1. | If the graph of \(f(x)\) is increasing, the graph of \(f'(x)\) is also increasing. |
Sometimes True
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| 2. | If the graph of \(f(x)\) has a local maximum, the graph of \(f'(x)\) has an \(x\)-intercept at the same \(x\) value. |
Always True
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| 3. | The graph of a quadratic function \(f(x)\) will have a gradient function \(f'(x)\) that is also a quadratic function. |
Never True
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| 4. | If \(f(x)\) is a straight line, \(f'(x)\) is a horizontal line. |
Always True
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| 5. | If the graph of \(f(x)\) is decreasing, the graph of \(f'(x)\) is below the \(x\)-axis. |
Always True
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| 6. | The \(y\)-intercept of the function \(f(x)\) is the same as the \(y\)-intercept of the gradient function \(f'(x)\). |
Sometimes True
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| 7. | Two functions, \(f(x)\) and \(g(x)\), that differ by a constant (e.g., \(f(x) = g(x) + 7\)) will have the same gradient function. |
Always True
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|
Notes
We can go further and actually find the gradient function by looking at the values of the gradients.
Consider the graph below of \(y=x^{2}\).
Find the gradient to \(y=x^{2}\) at each point given in the table.
You can use the interactive graph below to help find the gradient at each point more precisely.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-3\) |
\(-6\)
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|
| \(-2\) |
\(-4\)
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| \(-1\) |
\(-2\)
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| \(0\) |
\(0\)
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| \(1\) |
\(2\)
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| \(2\) |
\(4\)
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| \(3\) |
\(6\)
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|
What can we say about the gradient at the point \(k\) on the graph of \(y=x^{2}\)?
We can formalise this to say that if \(f\left(x\right)=x^{2}\) then \(f’\left(x\right)=2x\).
Recall that \(f’\left(x\right)\) is read as “\(f\) dashed of \(x\)” and is the notation for the derivative.
With this notation we can also have that \(f’\left(2\right)\) means the gradient of \(f\left(x\right)\) at \(x=2\).
So \(f’\left(2\right)=4\).
Your Turn
Draw the gradient function on the axes by first finding the gradient at each point given in the table.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-3\) |
\(-3\)
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|
| \(-2\) |
\(-1\)
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| \(-1\) |
\(1\)
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| \(0\) |
\(3\)
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| \(1\) |
\(5\)
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| \(2\) |
\(7\)
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| \(3\) |
\(9\)
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|
Your Turn
Draw the gradient function on the axes by first finding the gradient at each point given in the table.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-1\) |
\(7\)
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|
| \(0\) |
\(0\)
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|
| \(1\) |
\(-1\)
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| \(2\) |
\(4\)
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|
Your Turn
Draw the gradient function on the axes by first finding the gradient at each point given in the table.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-3\) |
\(-4.8\)
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|
| \(-2\) |
\(0.8\)
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|
| \(-1\) |
\(1.6\)
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| \(0\) |
\(0\)
?
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| \(1\) |
\(-1.6\)
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| \(2\) |
\(-0.8\)
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| \(3\) |
\(4.8\)
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|
Your Turn
Draw the gradient function on the axes by first finding the gradient at each point given in the table.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-3\) |
\(0.05\)
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|
| \(-2\) |
\(0.1\)
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| \(-1\) |
\(0.4\)
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|
| \(0\) |
\(1\)
?
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| \(1\) |
\(2.7\)
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| \(2\) |
\(7.4\)
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|
Your Turn
Draw the gradient function on the axes by first finding the gradient at each point given in the table.
| \(x\) | Gradient at \(x\) |
|---|---|
| \(-3\) |
\(-2\)
?
|
| \(-2\) |
\(-0.8\)
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| \(-1\) |
\(1.1\)
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| \(0\) |
\(2\)
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| \(1\) |
\(1.1\)
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| \(2\) |
\(-0.8\)
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| \(3\) |
\(-2\)
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Key Facts
Use this applet to generate a prompt for a Key Fact that you need to know for the course. The idea is that you should KNOW these key facts in order to be able to solve problems.
Taking it Deeper
Conceptual Questions to Consider
When you draw \(f’\left(x\right)\) you lose some information about the original function \(f\left(x\right)\). Why does the vertical position of the function not affect the gradient at a particular value of \(x\)? That is, why is the gradient function of \(y=x^{2}\) the same as for \(y=x^{2}+10\)?
Why is the gradient function of a straight line always going to be a horizontal line?
Why does a maximum point on \(f\left(x\right)\) map to a root on the graph of the derivative?
Common Mistakes / Misconceptions
A common mistake at this early stage is to confuse the notation \(f’\left(x\right)\) for \(f^{-1}\left(x\right)\). The first is the derivative, the second is the inverse function.
Connecting This to Other Skills
This skill builds on the idea of Tangent (4.1), and requires some basic knowledge of shapes of functions (2.7 - 2.10).
The idea of the derivative will be built upon formally from First Principles (4.4) and then we will see how we can work with derivatives algebraically when Differentiating Polynomials (4.5).
Finding Stationary Points (4.14) and Interpreting Derivatives (4.6) build on the link between this graphical perspective and the algebraic form.
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?