Unit 4 - Differential Calculus
Required Prior Knowledge
Questions
Simplify
a) \(\frac{x^{2}-9}{x+3} \)
b) \(\frac{x^{2}-5x+6}{x-2}\)
c) \(\frac{\frac{1}{x}}{x^{2}}\)
Solutions
Get Ready
Questions
Using the product rule, differentiate$$y=\frac{x^{2}+1}{x+1}$$
Solutions
Notes
The Quotient Rule for differentiation states that, of \(f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}\) then$$f’\left(x\right)=\frac{u’\left(x\right)v\left(x\right)-u\left(x\right)v’\left(x\right)}{\left(v\left(x\right)\right)^{2}}$$This can also be written as, if \(y=\frac{u}{v}\) where \(u\) and \(v\) are functions of \(x\), then$$\frac{dy}{dx}=\frac{u’v-uv’}{v^{2}}$$
Proof
Recall that for any function the derivative is defined from first principles as$$f’\left(x\right)=\lim_{h\to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}$$
Hence, for the function \(f\left(x\right)=\frac{u\left(x\right)}{v\left(x\right)}\) we have that:$$\begin{align}f’\left(x\right)&=\lim_{h\to 0}\frac{\frac{u\left(x+h\right)}{v\left(x+h\right)}-\frac{u\left(x\right)}{v\left(x\right)}}{h}\\&=\lim_{h\to 0}\frac{\frac{u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)}{v\left(x\right)v\left(x+h\right)}}{h}\\&=\lim_{h\to 0}\frac{u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)}{h v\left(x\right)v\left(x+h\right)}\\&=\lim_{h\to 0}\frac{u\left(x+h\right)v\left(x\right)-\left[u\left(x\right)v\left(x\right)-u\left(x\right)v\left(x\right)\right]-u\left(x\right)v\left(x+h\right)}{h v\left(x\right)v\left(x+h\right)}\\&=\lim_{h\to 0}\frac{u\left(x+h\right)v\left(x\right)-u\left(x\right)v\left(x\right)+u\left(x\right)v\left(x\right)-u\left(x\right)v\left(x+h\right)}{h v\left(x\right)v\left(x+h\right)}\\&=\lim_{h\to 0}\frac{v\left(x\right)\left(u\left(x+h\right)-u\left(x\right)\right)+u\left(x\right)\left(v\left(x\right)-v\left(x+h\right)\right)}{h v\left(x\right)v\left(x+h\right)}\\&=\lim_{h\to 0}\frac{v\left(x\right)\left(u\left(x+h\right)-u\left(x\right)\right)-u\left(x\right)\left(v\left(x+h\right)-v\left(x\right)\right)}{h v\left(x\right)v\left(x+h\right)}\\&=\lim_{h\to 0}\frac{\frac{v\left(x\right)\left(u\left(x+h\right)-u\left(x\right)\right)}{h}-\frac{u\left(x\right)\left(v\left(x+h\right)-v\left(x\right)\right)}{h}}{v\left(x\right)v\left(x+h\right)}\\&=\frac{\lim_{h\to 0}v\left(x\right)\lim_{h\to 0}\frac{u\left(x+h\right)-u\left(x\right)}{h}-\lim_{h\to 0}u\left(x\right)\lim_{h\to 0}\frac{v\left(x+h\right)-v\left(x\right)}{h}}{\lim_{h\to 0}v\left(x\right)v\left(x+h\right)}\\&=\frac{v\left(x\right)u’\left(x\right)-u\left(x\right)v’\left(x\right)}{\left(v\left(x\right)\right)^{2}}\end{align}$$
Examples and Your Turns
Example
Find \(f’\left(x\right)\) if $$f\left(x\right)=\frac{1+3x}{x^{2}+1}$$
Your Turn
Differentiate$$f\left(x\right)=\frac{x^{2}}{x+3}$$
Your Turn
Find \(f’\left(x\right)\) if $$f\left(x\right)=\frac{x^{2}+1}{x^{2}-1}$$
Your Turn
Find \(\frac{dy}{dx}\) if $$y=\frac{\sqrt{x}}{\left(1-2x\right)^{2}}$$
Differentiate the function using the Quotient Rule
Your Turn
Consider the function \(f\left(x\right)=\frac{ax^{2}}{x-1}\) where \(a\in\mathbb{R}, a\ne 1\).
a) Show that \(f’\left(x\right)=\frac{ax\left(x-2\right)}{\left(x-1\right)^{2}}\).
b) The curve has a stationary point at \(\left(2,12\right)\). Find the value of \(a\).
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
Do you prefer to use the Quotient Rule or rearrange and use the Product Rule? Why?
Can you prove the Quotient Rule for \(y=\frac{u}{v}\) by rewriting it as \(y=uv^{-1}\) and using the Product Rule?
What is the key difference between the formula for the Product Rule and the Quotient Rule?
Why do I teach the ‘cross’ starting in the bottom left for the Product Rule?
Common Mistakes / Misconceptions
The most common mistake is to do the cross in the wrong order. Whilst it does not matter for the Product Rule, the subtraction means the order is important for the Quotient Rule.
Another common error is to either forget the denominator of \(v^{2}\) entirely, or to use \(v\) or \(v’\) for the denominator.
Both these errors are avoidable if you check the formula in the formula booklet.
The final common error is in the cancellation and simplification of the resulting fraction. Factorise the numerator first, simplify and then look for factors to cancel.
Connecting This to Other Skills
The Quotient Rule is related to the Product Rule (4.8) and requires knowledge of Differentiating Polynomials (4.5) and the Chain Rule (4.7).
The simplification of the fraction obtained requires a good knowledge of Algebraic Fractions (PK5).
You might be asked to Interpret Derivatives (4.6), find Tangents and Normals (4.13) or Stationary Points (4.14) where the Quotient Rule is needed.
Self-Reflection
What was the most challenging part of this skill for you?
What are you still unsure about that you need to review?