S4 Mathematics Practice Book · Unit 8: Differentiation of polynomial, rational and irrational functions

Unit 8 Differentiation of polynomial, rational and irrational functions

Key unit competence: Apply derivatives of polynomial, rational and irrational functions in solving related mathematical and real-life problems.

Key ideas

The derivative of $f$ at $x$ is $f^{'} (x) = lim_{h \to 0} \frac{f ( x + h ) - f ( x )}{h}$ : the gradient of the tangent, the instantaneous rate of change.
Rules: $(c)^{'} = 0$ ; $(x^{n})^{'} = n x^{n - 1}$ ; $(c f)^{'} = c f^{'}$ ; $(f \pm g)^{'} = f^{'} \pm g^{'}$ .
Product rule $(uv)^{'} = u^{'} v + u v^{'}$ ; quotient rule $(\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}$ ; chain rule $[f (g (x))]^{'} = f^{'} (g (x)) g^{'} (x)$ ; $(u)^{'} = \frac{u ^{'}}{2 u}$ .
Tangent at $x = a$ : $y - f (a) = f^{'} (a) (x - a)$ . Normal: gradient $- \frac{1}{f ^{'} ( a )}$ .
Higher derivatives: $f^{''} = (f^{'})^{'}$ . Velocity $v = \frac{d s}{d t}$ , acceleration $a = \frac{d v}{d t} = \frac{d ^{2} s}{d t ^{2}}$ .
Applications: $f^{'} > 0$ : increasing; $f^{'} < 0$ : decreasing; $f^{'} (a) = 0$ : stationary point; $f^{''} (a) > 0$ minimum, $f^{''} (a) < 0$ maximum. Optimisation problems: write the quantity as a function of one variable, differentiate, set to 0.

Formulas to know

f^{'} (x) = h \to 0 lim \frac{f ( x + h ) - f ( x )}{h}

(x^{n})^{'} = n x^{n - 1}

(uv)^{'} = u^{'} v + u v^{'}, (\frac{u}{v})^{'} = \frac{u ^{'} v - u v ^{'}}{v ^{2}}

(f (g (x)))^{'} = f^{'} (g (x)) g^{'} (x)

Common misconceptions

" $(uv)^{'} = u^{'} v^{'}$ ." No: use the product rule $u^{'} v + u v^{'}$ .
" $\frac{d}{d x} (x^{3} + 5)^{2} = 2 (x^{3} + 5)$ ." No: by the chain rule, multiply by the inner derivative $3 x^{2}$ : $6 x^{2} (x^{3} + 5)$ .
" $f^{'} (a) = 0$ always means a maximum or minimum." No: it can be a point of inflection (e.g. $x^{3}$ at 0).

Worked examples

Example 1. Differentiate $f (x) = \frac{x ^{2} + 1}{x - 1}$ .

Solution

$f^{'} (x) = \frac{2 x ( x - 1 ) - ( x ^{2} + 1 )}{( x - 1 ) ^{2}} = \frac{x ^{2} - 2 x - 1}{( x - 1 ) ^{2}}$ .

Example 2. Find the tangent to $y = x^{2} - 3 x$ at $x = 2$ .

Solution

$y (2) = - 2$ , $y^{'} = 2 x - 3$ , $y^{'} (2) = 1$ . Tangent: $y + 2 = x - 2$ , i.e. $y = x - 4$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

Use the definition to find the derivative of $f (x) = 3 x^{2}$ .
Show answer
$\frac{3 ( x + h ) ^{2} - 3 x ^{2}}{h} = \frac{6 x h + 3 h ^{2}}{h} = 6 x + 3 h \to 6 x$ .
Common mistake: Forgetting to divide by $h$ before letting $h \to 0$ .
Find the points on $y = x^{3} - 3 x$ where the tangent is horizontal.
Show answer
$3 x^{2} - 3 = 0$ , $x = \pm 1$ : $(1, - 2)$ and $(- 1, 2)$ .
Common mistake: Forgetting the negative solution.
Differentiate $y = (x^{2} - 3 x)^{4}$ .
Show answer
$4 (x^{2} - 3 x)^{3} (2 x - 3)$ .
Common mistake: Leaving out the factor $(2 x - 3)$ .
A stone is thrown up with height $h = 30 t - 5 t^{2}$ . Find its velocity at $t = 1$ , the time of maximum height and the maximum height.
Show answer
$v = 30 - 10 t$ ; $v (1) = 20$ m/s; $v = 0$ at $t = 3$ s; $h (3) = 45$ m.
Common mistake: Confusing velocity with height.
Differentiate $y = \frac{2 x - 1}{x ^{2} + 1}$ .
Show answer
$\frac{2 ( x ^{2} + 1 ) - ( 2 x - 1 ) 2 x}{( x ^{2} + 1 ) ^{2}} = \frac{- 2 x ^{2} + 2 x + 2}{( x ^{2} + 1 ) ^{2}}$ .
Common mistake: Reversing the order in the numerator of the quotient rule.
The cost of producing $x$ items is $C = x^{2} - 40 x + 600$ thousand FRW. How many items minimise the cost, and what is the minimum?
Show answer
$C^{'} = 2 x - 40 = 0$ , $x = 20$ ; $C = 400 - 800 + 600 = 200$ thousand FRW.
Common mistake: Setting $C = 0$ instead of $C^{'} = 0$ .
Find the equations of the tangent and normal to $y = \frac{4}{x}$ at $x = 2$ .
Show answer
Point $(2, 2)$ ; $y^{'} = - \frac{4}{x ^{2}} = - 1$ . Tangent $y = - x + 4$ ; normal gradient 1: $y = x$ .
Common mistake: Using the same gradient for both.
Find and classify the stationary points of $y = 2 x^{3} - 3 x^{2} - 12 x$ .
Show answer
$y^{'} = 6 (x^{2} - x - 2) = 6 (x - 2) (x + 1)$ . $x = - 1$ : $y = 7$ , max ( $y^{''} = - 18$ ). $x = 2$ : $y = - 20$ , min ( $y^{''} = 18$ ).
Common mistake: Classifying without the second derivative or a sign table.

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. $y = \frac{x}{x ^{2} + 1}$ . Find $y^{'} (0)$ .	a. $1$
2. Find the normal to $y = x^{2}$ at $(1, 1)$ .	b. $- \frac{4}{x ^{3}} + \frac{3}{2 x}$
3. Find the minimum value of $f (x) = x + \frac{4}{x}$ for $x > 0$ .	c. $y = - \frac{1}{2} x + \frac{3}{2}$
4. Differentiate $y = \frac{2}{x ^{2}} + 3 x$ .	d. $4$

Show answer

1 → a ( $1$ ) · 2 → c ( $y = - \frac{1}{2} x + \frac{3}{2}$ ) · 3 → d ( $4$ ) · 4 → b ( $- \frac{4}{x ^{3}} + \frac{3}{2 x}$ )

Step 1. $y^{'} = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}}$ .
Step 2. $y^{'} (0) = \frac{1}{1} = 1$ .
Step 1. Tangent gradient $2 x = 2$ .
Step 2. Normal gradient $- \frac{1}{2}$ : $y - 1 = - \frac{1}{2} (x - 1)$ .
Step 3. $y = - \frac{1}{2} x + \frac{3}{2}$ .
Step 1. $f^{'} (x) = 1 - \frac{4}{x ^{2}} = 0$ , so $x = 2$ .
Step 2. $f (2) = 2 + 2 = 4$ ; $f^{''} (2) = \frac{8}{8} > 0$ : minimum.
Step 1. Write as powers: $2 x^{- 2} + 3 x^{\frac{1}{2}}$ .
Step 2. $- 4 x^{- 3} + \frac{3}{2} x^{- \frac{1}{2}}$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$x = 2$ cm · $6 x + 3 h$ · $x = 3$ cm · $100$ cm²
1. An open box is made from a 12 cm square sheet by cutting squares of side $x$ from the corners. The volume is maximum when .
2. For $f (x) = 3 x^{2}$ , the difference quotient $\frac{f ( x + h ) - f ( x )}{h}$ simplifies to .
3. A rectangle has perimeter 40 cm. Its maximum area is .
Show answer
1: $x = 2$ cm · 2: $6 x + 3 h$ · 3: $100$ cm²
- Step 1. $V = x (12 - 2 x)^{2}$ .
  Step 2. $V^{'} = (12 - 2 x) (12 - 6 x) = 0$ : $x = 2$ or $x = 6$ .
  Step 3. $x = 6$ gives no box; $V (2) = 128$ cm³ is the maximum.
- Step 1. $f (x + h) - f (x) = 3 x^{2} + 6 x h + 3 h^{2} - 3 x^{2} = 6 x h + 3 h^{2}$ .
  Step 2. Divide by $h$ : $6 x + 3 h$ (its limit as $h \to 0$ is $6 x$ ).
- Step 1. Sides $x$ and $20 - x$ : $A = 20 x - x^{2}$ .
  Step 2. $A^{'} = 20 - 2 x = 0$ , $x = 10$ : a square, $A = 100$ cm².

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. Find the point of inflection of $y = x^{3} - 3 x^{2} + 2$ .	a. $16 x (2 x^{2} - 3)^{3}$
2. Differentiate $f (x) = \frac{3 x + 2}{x - 1}$ .	b. $y = x - 2$
3. Differentiate $f (x) = (2 x^{2} - 3)^{4}$ .	c. $- \frac{5}{( x - 1 ) ^{2}}$
4. Find the tangent to $y = x^{3} - 2 x$ at $x = 1$ .	d. $(1, 0)$

Show answer

1 → d ( $(1, 0)$ ) · 2 → c ( $- \frac{5}{( x - 1 ) ^{2}}$ ) · 3 → a ( $16 x (2 x^{2} - 3)^{3}$ ) · 4 → b ( $y = x - 2$ )

Step 1. $y^{''} = 6 x - 6 = 0$ at $x = 1$ , and $y^{''}$ changes sign there.
Step 2. $y (1) = 1 - 3 + 2 = 0$ .
Step 1. Quotient rule: $\frac{3 ( x - 1 ) - ( 3 x + 2 ) ( 1 )}{( x - 1 ) ^{2}}$ .
Step 2. Numerator: $3 x - 3 - 3 x - 2 = - 5$ .
Step 1. Chain rule: $4 (2 x^{2} - 3)^{3} \times (2 x^{2} - 3)^{'}$ .
Step 2. $(2 x^{2} - 3)^{'} = 4 x$ , so $16 x (2 x^{2} - 3)^{3}$ .
Step 1. Point: $y (1) = 1 - 2 = - 1$ , so $(1, - 1)$ .
Step 2. Gradient: $y^{'} = 3 x^{2} - 2 = 1$ .
Step 3. $y + 1 = 1 (x - 1)$ , so $y = x - 2$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$100$ · $200$ · $(2, - 11)$ · $x < 1$ or $x > 3$
1. Two positive numbers have sum 20. Their largest possible product is .
2. On which intervals is $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ increasing? →
3. Find the point of inflection of $y = x^{3} - 6 x^{2} + 5$ . →
Show answer
1: $100$ · 2: $x < 1$ or $x > 3$ · 3: $(2, - 11)$
- Step 1. Product $P = x (20 - x)$ , so $P^{'} = 20 - 2 x$ .
  Step 2. $P^{'} = 0$ at $x = 10$ ; $P^{''} = - 2 < 0$ : maximum $10 \times 10 = 100$ .
- Step 1. $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ .
  Step 2. $f^{'} > 0$ outside the roots: $x < 1$ or $x > 3$ .
- Step 1. $y^{''} = 6 x - 12 = 0$ at $x = 2$ ; it changes sign there.
  Step 2. $y (2) = 8 - 24 + 5 = - 11$ .

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