➗ Derivative of a Quotient – Quotient Rule
How to differentiate a rational function (fraction)
🎯 Why Does This Matter?
To investigate a rational function (find increasing/decreasing intervals, extrema, concavity) – we need to differentiate it!
Quotient functions have a special differentiation rule – the Quotient Rule.
⚠️ Note: you cannot differentiate numerator and denominator separately! You must use the formula.
⭐ Quotient Rule – The Formula
If \(f(x) = \frac{u(x)}{v(x)}\) then:
\(f'(x) = \frac{u' \cdot v - u \cdot v'}{v^2}\)
where \(u\) = numerator, \(v\) = denominator
🎵 How to remember? – "The Quotient Song"
"Derivative of numerator times denominator
minus numerator times derivative of denominator
all over denominator squared"
\(\frac{u' \cdot v - u \cdot v'}{v^2}\)
⚠️ Order matters!
It is \(u'v - uv'\) and not \(uv' - u'v\)
The order is important! (unlike the product rule)
✏️ Example 1 – Simple Start
Find the derivative of: \(f(x) = \frac{x+1}{x-2}\)
Step 1: Identify \(u\) and \(v\)
\(u = x + 1\) \(v = x - 2\)
Step 2: Differentiate each separately
\(u' = 1\) \(v' = 1\)
Step 3: Substitute into the formula
\(f'(x) = \frac{u' \cdot v - u \cdot v'}{v^2}\)
\(f'(x) = \frac{1 \cdot (x-2) - (x+1) \cdot 1}{(x-2)^2}\)
Step 4: Simplify the numerator
\(f'(x) = \frac{x - 2 - x - 1}{(x-2)^2}\)
\(f'(x) = \frac{-3}{(x-2)^2}\)
\(f'(x) = \frac{-3}{(x-2)^2}\)
✏️ Example 2 – With Squares
Find the derivative of: \(f(x) = \frac{x^2}{x+3}\)
Step 1: Identify
\(u = x^2\) \(v = x + 3\)
Step 2: Differentiate
\(u' = 2x\) \(v' = 1\)
Step 3: Substitute
\(f'(x) = \frac{2x \cdot (x+3) - x^2 \cdot 1}{(x+3)^2}\)
Step 4: Simplify
\(f'(x) = \frac{2x^2 + 6x - x^2}{(x+3)^2}\)
\(f'(x) = \frac{x^2 + 6x}{(x+3)^2}\)
\(f'(x) = \frac{x^2 + 6x}{(x+3)^2}\)
or after factoring: \(f'(x) = \frac{x(x + 6)}{(x+3)^2}\)
✏️ Example 3 – Quadratic Denominator
Find the derivative of: \(f(x) = \frac{2x+1}{x^2-4}\)
Step 1: Identify
\(u = 2x + 1\) \(v = x^2 - 4\)
Step 2: Differentiate
\(u' = 2\) \(v' = 2x\)
Step 3: Substitute
\(f'(x) = \frac{2 \cdot (x^2-4) - (2x+1) \cdot 2x}{(x^2-4)^2}\)
Step 4: Simplify the numerator
\(= \frac{2x^2 - 8 - 4x^2 - 2x}{(x^2-4)^2}\)
\(= \frac{-2x^2 - 2x - 8}{(x^2-4)^2}\)
\(f'(x) = \frac{-2x^2 - 2x - 8}{(x^2-4)^2}\)
or: \(f'(x) = \frac{-2(x^2 + x + 4)}{(x^2-4)^2}\)
🔢 Special Case: Constant Numerator
When \(f(x) = \frac{k}{v(x)}\) (the numerator is a constant), the formula simplifies:
\(\left(\frac{k}{v}\right)' = \frac{-k \cdot v'}{v^2}\)
Why? Because \(u = k\) so \(u' = 0\), therefore:
\(\frac{0 \cdot v - k \cdot v'}{v^2} = \frac{-k \cdot v'}{v^2}\)
Example: \(f(x) = \frac{1}{x}\)
\(k = 1\), \(v = x\), \(v' = 1\)
\(f'(x) = \frac{-1 \cdot 1}{x^2} = -\frac{1}{x^2}\)
Example: \(f(x) = \frac{3}{x^2+1}\)
\(k = 3\), \(v = x^2+1\), \(v' = 2x\)
\(f'(x) = \frac{-3 \cdot 2x}{(x^2+1)^2} = \frac{-6x}{(x^2+1)^2}\)
📊 Useful Derivatives to Memorise
| Function \(f(x)\) | Derivative \(f'(x)\) |
|---|---|
| \(\frac{1}{x}\) | \(-\frac{1}{x^2}\) |
| \(\frac{1}{x^2}\) | \(-\frac{2}{x^3}\) |
| \(\frac{1}{x+a}\) | \(-\frac{1}{(x+a)^2}\) |
| \(\frac{x}{x+a}\) | \(\frac{a}{(x+a)^2}\) |
| \(\frac{ax+b}{cx+d}\) | \(\frac{ad-bc}{(cx+d)^2}\) |
💡 Tip: It is worth memorising the common derivatives to save time in the exam!
💡 Important Tips for the Exam
1️⃣ Order matters!
\(u'v - uv'\) ✓
\(uv' - u'v\) ✗
Always derivative of numerator first!
2️⃣ Don't forget to simplify!
After substituting – expand the brackets
and collect like terms in the numerator
3️⃣ Denominator squared!
Always \(v^2\) in the denominator
Do not expand the square (leave it)
4️⃣ Organise your work
Write on the side:
\(u = ...\) \(u' = ...\)
\(v = ...\) \(v' = ...\)
⚠️ Common Mistakes
| ❌ Mistake | ✓ Correct |
|---|---|
| \(\left(\frac{u}{v}\right)' = \frac{u'}{v'}\) | \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\) |
| \(\frac{uv' - u'v}{v^2}\) (wrong order) | \(\frac{u'v - uv'}{v^2}\) |
| \(\frac{u'v - uv'}{v}\) (forgot the square) | \(\frac{u'v - uv'}{v^2}\) |
📝 Summary
\(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\)
"Derivative of numerator times denominator, minus numerator times derivative of denominator, all over denominator squared"
Now you are ready for the summary page: Investigating a Rational Function!