Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
a) Write \(\frac{4}{x^{2}}-\sqrt{x} \) in the form \(ax^{n}+bx^{m}\)
b) Expand and simplify \(\left(2x-3\right)^{2}\)
Solutions
Get Ready (HL - only this part)
Questions
From first principles, find the derivative of the function \(y=ax^{n}\).
Solutions
Notes
We can summarise the rules of differentiating polynomials in the following table:
| \( f(x) \) | \( f'(x) \) | Name of rule |
|---|---|---|
| \( x^n \) | \( n x^{n-1} \) | Power rule |
| \( c \) | \( 0 \) | Constants |
| \( k f(x) \) | \( k f'(x) \) | Scalar multiples |
| \( f(x) + g(x) \) | \( f'(x) + g'(x) \) | Addition rule |
Examples and Your Turns
Example
Find the derivative for:
a) \(y=x^{3}\)
b) \(f\left(x\right)=4x^{2}\)
c) \(f\left(x\right)=\sqrt{x}\)
d) \(y=\frac{1}{x}\)
Your Turn
Differentiate \(y=x^{5}-4x^{3}+7\)
Your Turn
Find the derivative of \(f\left(x\right)=\left(x+2\right)\left(2x-3\right)\)
Your Turn
If \(y=x^{2}\sqrt{x}\) find \(\frac{dy}{dx}\)
Your Turn
Find \(f’\left(x\right)\) given that \(f\left(x\right)=\frac{2\left(2x-7\right)}{\sqrt{x}}\)
Your Turn
If \(y=\frac{1}{2x^{3}}+\frac{1}{2}\) find \(\frac{dy}{dx}\)
Your Turn
Differentiate \(f\left(x\right)=\frac{4}{5\sqrt{x^{3}}}\)
Practice
Differentiate each function below.
| Function (\(y\)) | Prepared for Differentiation | \(\frac{dy}{dx}\) | Tidied Up |
|---|---|---|---|
| \(y = x^5\) | \(y = x^5\) |
\(5x^4\)
?
|
\(5x^4\)
?
|
| \(y = \frac{1}{x^3}\) |
\(y = x^{-3}\)
?
|
\(-3x^{-4}\)
?
|
\(-\frac{3}{x^4}\)
?
|
| \(y = \sqrt{x}\) |
\(y = x^{1/2}\)
?
|
\(\frac{1}{2}x^{-1/2}\)
?
|
\(\frac{1}{2\sqrt{x}}\)
?
|
| \(y = \frac{3}{x^2}\) |
\(y = 3x^{-2}\)
?
|
\(-6x^{-3}\)
?
|
\(-\frac{6}{x^3}\)
?
|
| \(y = \sqrt[3]{x}\) |
\(y = x^{1/3}\)
?
|
\(\frac{1}{3}x^{-2/3}\)
?
|
\(\frac{1}{3\sqrt[3]{x^2}}\)
?
|
| \(y = \frac{x^4}{2}\) |
\(y = \frac{1}{2}x^4\)
?
|
\(2x^3\)
?
|
\(2x^3\)
?
|
| \(y = \frac{1}{\sqrt{x}}\) |
\(y = x^{-1/2}\)
?
|
\(-\frac{1}{2}x^{-3/2}\)
?
|
\(-\frac{1}{2\sqrt{x^3}}\)
?
|
| \(y = \sqrt{x^5}\) |
\(y = x^{5/2}\)
?
|
\(\frac{5}{2}x^{3/2}\)
?
|
\(\frac{5\sqrt{x^3}}{2}\)
?
|
| \(y = \frac{1}{4x^2}\) |
\(y = \frac{1}{4}x^{-2}\)
?
|
\(-\frac{1}{2}x^{-3}\)
?
|
\(-\frac{1}{2x^3}\)
?
|
| \(y = 3\sqrt{x} + \frac{2}{x}\) |
\(y = 3x^{1/2} + 2x^{-1}\)
?
|
\(\frac{3}{2}x^{-1/2} - 2x^{-2}\)
?
|
\(\frac{3}{2\sqrt{x}} - \frac{2}{x^2}\)
?
|
| \(y = \sqrt[4]{x}\) |
\(y = x^{1/4}\)
?
|
\(\frac{1}{4}x^{-3/4}\)
?
|
\(\frac{1}{4\sqrt[4]{x^3}}\)
?
|
| \(y = \frac{10}{\sqrt{x}}\) |
\(y = 10x^{-1/2}\)
?
|
\(-5x^{-3/2}\)
?
|
\(-\frac{5}{\sqrt{x^3}}\)
?
|
| \(y = \frac{x^2+3}{x}\) |
\(y = x + 3x^{-1}\)
?
|
\(1 - 3x^{-2}\)
?
|
\(1 - \frac{3}{x^2}\)
?
|
| \(y = \frac{4}{3x^2}\) |
\(y = \frac{4}{3}x^{-2}\)
?
|
\(-\frac{8}{3}x^{-3}\)
?
|
\(-\frac{8}{3x^3}\)
?
|
| \(y = \sqrt[3]{x^2}\) |
\(y = x^{2/3}\)
?
|
\(\frac{2}{3}x^{-1/3}\)
?
|
\(\frac{2}{3\sqrt[3]{x}}\)
?
|
| \(y = \frac{1}{2x\sqrt{x}}\) |
\(y = \frac{1}{2}x^{-3/2}\)
?
|
\(-\frac{3}{4}x^{-5/2}\)
?
|
\(-\frac{3}{4\sqrt{x^5}}\)
?
|
| \(y = \frac{5}{\sqrt[3]{x}}\) |
\(y = 5x^{-1/3}\)
?
|
\(-\frac{5}{3}x^{-4/3}\)
?
|
\(-\frac{5}{3\sqrt[3]{x^4}}\)
?
|
| \(y = \frac{\sqrt{x}}{4}\) |
\(y = \frac{1}{4}x^{1/2}\)
?
|
\(\frac{1}{8}x^{-1/2}\)
?
|
\(\frac{1}{8\sqrt{x}}\)
?
|
| \(y = \frac{1}{x^{10}}\) |
\(y = x^{-10}\)
?
|
\(-10x^{-11}\)
?
|
\(-\frac{10}{x^{11}}\)
?
|
| \(y = \frac{3}{2\sqrt[3]{x^2}}\) |
\(y = \frac{3}{2}x^{-2/3}\)
?
|
\(-x^{-5/3}\)
?
|
\(-\frac{1}{\sqrt[3]{x^5}}\)
?
|
Differentiate the function (Rewrite, Differentiate, Simplify)
1. Original Function
2. Rewrite in Index Form
3. Differentiate using Power Rule
4. Simplify to Standard Form
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
Why is the derivative of \(y=5x\) equal to \(5\)? Think about the graph.
Why is the derivative of a constant \(y=c\) equal to \(0\)? Think about the graph.
Common Mistakes / Misconceptions
A common mistake is to forget that constant differentiate to \(0\).
Be careful with fractions and correctly writing them in the form \(ax^{n}\). For example, \(\frac{1}{2x^{3}}\) is NOT \(2x^{-3}\) as the \(2\) is not part of the base of the power. The correct simplification is \(\frac{1}{2x^{3}}=\frac{1}{2}x^{-3}\).
Over-generalising the sum rule. It IS true that the derivative of \(f\left(x\right)+g\left(x\right)\) is \(f’\left(x\right)+g’\left(x\right)\). But the same is NOT true for multiplication. That is the derivative of \(f\left(x\right)g\left(x\right)\) is NOT \(f’\left(x\right)g’\left(x\right)\). We can see this by considering the function \(f\left(x\right)=\left(x+2\right)\left(x+3\right)\). This expands to the quadratic \(f\left(x\right)=x^{2}+5x+6\) and so \(f’\left(x\right)=2x+5\). It is not equal to \(1\), which is what we would get if we differentiate each bracket and multiply. In order to differentiate products we need the Product Rule (4.8).
Connecting This to Other Skills
We have built upon the idea of a Gradient Function (4.3) and used First Principles (4.4) to show this result for Polynomials. This is the most basic rule in differentiation.
We will build upon this with more rules: Chain Rule (4.7), Product Rule (4.8) and Quotient Rule (4.9). For multivariable functions we see Implicit Differentiation (4.10).
Then we see other types of functions such as Differentiating Exponentials and Logarithms (4.11) and Differentiating Trig (4.12).
We also see the applications of these ideas when Interpreting Derivatives (4.6), finding Tangents and Normals (4.13) and Stationary Points (4.14).
When we get to unit 5, we see Anti-derivatives (5.1) and how we can reverse this process.
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?