Differentiating Polynomials

Finding dy/dx with the power rule, tidying fractions and brackets first, where the gradient is a given number, finding unknown constants, and velocity and acceleration (Higher).

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1 · 29 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

What dy/dx Means (Higher)

A curve is steep in some places and flat in others. $\frac{d y}{d x}$ is a formula for the gradient at any x.

Example

On a hill, the gradient tells you how steep the path is where you stand. Put in your x and you get the steepness there.

Rule

Finding $\frac{d y}{d x}$ is called differentiating. The 2-mark “Find dy/dx” is the most common question here.

Key idea

Some questions write the curve as $f (x)$ . Then its gradient formula is written $f^{'} (x)$ . It means the same as $\frac{d y}{d x}$ .

Key idea

Calculus was invented twice: by Isaac Newton in England and by the German Gottfried Leibniz, working separately in the late 1600s. The $\frac{d y}{d x}$ you write today is Leibniz's notation. It first appears in his private notes around 1675.

The Power Rule

Each term is done on its own. Multiply by the power, then take 1 off the power.

Formula

\frac{d}{d x} (x^{n}) = n x^{n - 1}

Here $n$ is the power of $x$ , and $\frac{d}{d x}$ means “differentiate with respect to $x$ ”. If there is a number in front, it waits and multiplies the answer:

Formula

\frac{d}{d x} (a x^{n}) = an x^{n - 1}

Worked example

1
$y = x^{3} - 2 x^{2} - 15 x + 5$ — Four terms. Differentiate each one on its own, then put the results together.
2
$x^{3} \to 3 x^{2}$ — The power 3 comes down in front, and the power drops by 1: 3 − 1 = 2.
3
$- 2 x^{2} \to - 4 x$ — The power 2 multiplies the −2: 2 × (−2) = −4. The power drops from 2 to 1, and $x^{1}$ is just $x$ .
4
$- 15 x \to - 15$ — $x$ is $x^{1}$ , so 1 × (−15) = −15. The power drops to 0 and $x^{0} = 1$ , so the $x$ disappears.
5
$5 \to 0$ — A number on its own never changes as $x$ changes, so its gradient is 0.
6
$\frac{d y}{d x} = 3 x^{2} - 4 x - 15$ — Put the four results together.

Common mistake

Keeping the number on its own. A flat number has no slope, so it becomes 0.

Key idea

Show each term. Examiners give a mark for getting some terms right, so one slip does not lose everything.

Tidy It First

The rule only works on single powers of x. Brackets and fractions must be rewritten first.

Worked example

1
$y = 4 x^{2} + \frac{125}{x}$ — The second term has $x$ on the bottom, so the power rule cannot be used yet.
2
$\frac{125}{x} = 125 x^{- 1}$ — Moving $x$ from the bottom to the top makes its power negative.
3
$y = 4 x^{2} + 125 x^{- 1}$ — Now every term is a number times a power of $x$ .
4
$4 x^{2} \to 8 x$ — 2 × 4 = 8, and the power drops from 2 to 1.
5
$125 x^{- 1} \to - 125 x^{- 2}$ — The power −1 multiplies 125 to give −125. Take 1 off the power: −1 − 1 = −2.
6
$\frac{d y}{d x} = 8 x - \frac{125}{x ^{2}}$ — Put the terms together and write $x^{- 2}$ back as a fraction.

Worked example

1
$y = (2 x + 1) (x - 3)$ — Two brackets multiplied together, so the power rule cannot be used yet.
2
$y = 2 x^{2} - 6 x + x - 3$ — Expand: $2 x \times x = 2 x^{2}$ , $2 x \times (- 3) = - 6 x$ , $1 \times x = x$ and $1 \times (- 3) = - 3$ .
3
$y = 2 x^{2} - 5 x - 3$ — Collect the $x$ terms: −6x + x = −5x.
4
$\frac{d y}{d x} = 4 x - 5$ — Power rule on each term: $2 x^{2} \to 4 x$ , $- 5 x \to - 5$ and $- 3 \to 0$ .

Common mistake

Differentiating (2x + 1)(x − 3) one bracket at a time. Expand first: $2 x^{2} - 5 x - 3$ gives $4 x - 5$ .

Key idea

A fraction with one term underneath, like $\frac{3 x ^{3} + 2 x ^{2}}{x}$ : divide every term by $x$ first, to get $3 x^{2} + 2 x$ .

Where the Gradient Is a Number

“Find the points where the gradient is 8”: set $\frac{d y}{d x}$ equal to 8 and solve. Then find each $y$ .

Worked example

1
$y = x^{3} - 40 x + 1$ — The curve. We want the points on it where the gradient is 8.
2
$\frac{d y}{d x} = 3 x^{2} - 40$ — Differentiate to get the gradient formula.
3
$3 x^{2} - 40 = 8$ — The gradient must be 8, so set the formula equal to 8.
4
$3 x^{2} = 48$ — Add 40 to both sides.
5
$x^{2} = 16$ — Divide both sides by 3.
6
$x = 4 or x = - 4$ — Square root both sides. A square root has a plus and a minus answer, so there are two points.
7
$y = 4^{3} - 40 (4) + 1 = - 95$ — These are $x$ values. Put each into the curve, not into $\frac{d y}{d x}$ , to get $y$ : 64 − 160 + 1 = −95.
8
$y = (- 4)^{3} - 40 (- 4) + 1 = 97$ — Use brackets for −4: $(- 4)^{3} = - 64$ and $- 40 \times (- 4) = + 160$ , so −64 + 160 + 1 = 97.
9
$(4, - 95) and (- 4, 97)$ — Write each point with $x$ first, then $y$ .

Worked example

1
$y = 2 x^{3} + 3 x^{2} - 11 x + 1$ — The gradient is 1 at two points. The curve is given.
2
$\frac{d y}{d x} = 6 x^{2} + 6 x - 11$ — Differentiate each term.
3
$6 x^{2} + 6 x - 11 = 1$ — Set the gradient equal to 1.
4
$6 x^{2} + 6 x - 12 = 0$ — Take 1 from both sides, so one side is 0.
5
$x^{2} + x - 2 = 0$ — Divide every term by 6.
6
$(x + 2) (x - 1) = 0$ — Factorise: two numbers that multiply to −2 and add to 1 are 2 and −1.
7
$x = - 2 or x = 1$ — Each bracket equal to 0 gives one answer.
8
$(- 2, 19) and (1, - 5)$ — Put each $x$ into the curve: $2 (- 8) + 3 (4) + 22 + 1 = 19$ and $2 + 3 - 11 + 1 = - 5$ .

Common mistake

Putting x back into $\frac{d y}{d x}$ instead of y. The point is on the curve, so use y.

Key idea

If the quadratic will not factorise, use the quadratic formula. Either way, get 0 on one side first.

Finding Unknown Numbers (Higher)

The 6-mark version gives a point and a gradient. Each fact makes one equation.

Worked example

1
$y = a x^{3} + b x^{2} - 12 x + 6$ — A(2, −6) is on the curve, and the gradient at A is 16. $a$ and $b$ are unknown numbers.
2
$- 6 = a (2)^{3} + b (2)^{2} - 12 (2) + 6$ — Fact 1, the point: put $x = 2$ and $y = - 6$ into the curve.
3
$8 a + 4 b - 24 + 6 = - 6$ — Work out the powers: $2^{3} = 8$ , $2^{2} = 4$ and 12 × 2 = 24.
4
$8 a + 4 b = 12$ — −24 + 6 = −18. Add 18 to both sides: −6 + 18 = 12.
5
$2 a + b = 3$ — Divide every term by 4 to keep the numbers small.
6
$\frac{d y}{d x} = 3 a x^{2} + 2 b x - 12$ — Fact 2, the gradient: differentiate. $a$ and $b$ are just numbers, so they stay in front.
7
$12 a + 4 b - 12 = 16$ — Put $x = 2$ into the gradient formula and set it equal to 16.
8
$3 a + b = 7$ — Add 12 to both sides to get $12 a + 4 b = 28$ , then divide by 4.
9
$(3 a + b) - (2 a + b) = 7 - 3$ — Subtract the first equation from the second. The $b$ cancels.
10
$a = 4$ — 3a − 2a = a and 7 − 3 = 4.
11
$b = 3 - 2 (4) = - 5$ — Put $a = 4$ into $2 a + b = 3$ : $b = 3 - 8$ .
12
$8 (4) + 4 (- 5) = 12$ — Check in the first equation: 32 − 20 = 12. It works.

Key idea

Two unknowns need two equations. The point gives one and the gradient gives the other.

Matching a Given Derivative (Higher)

Sometimes you are given dy/dx and must find missing numbers in $y$ . Differentiate $y$ with the letters in, then match it to the given gradient formula, term by term.

Worked example

1
$y = x^{n} + p x^{2} + 8 x, \frac{d y}{d x} = 4 x^{3} - 6 x + 8$ — The curve has two unknown numbers, $n$ and $p$ . We are told what $\frac{d y}{d x}$ is.
2
$\frac{d y}{d x} = n x^{n - 1} + 2 p x + 8$ — Differentiate each term. The power comes down and drops by 1, and the letters stay.
3
$n x^{n - 1} = 4 x^{3}$ — Match the $x^{3}$ terms. Only the $x^{n}$ term can make an $x^{3}$ term.
4
$n - 1 = 3 \Rightarrow n = 4$ — The powers must be equal. Check the number in front: $n = 4$ gives 4, which matches.
5
$2 p = - 6$ — Match the $x$ terms. The numbers in front must be equal.
6
$p = - 6 \div 2 = - 3$ — Divide both sides by 2. Keep the minus sign.
7
$8 x \to 8$ — Check the last term: $8 x$ gives 8, the same as the question, so nothing more is needed.

Key idea

Letters in front work the same way. The derivative of $2 a x^{7}$ is $14 a x^{6}$ . If that is $42 x^{6}$ , then $14 a = 42$ and $a = 3$ .

Speed and Acceleration (Higher)

$s$ is the distance from a fixed point at time $t$ . Differentiate once for velocity, twice for acceleration.

Formula

v = \frac{d s}{d t}, a = \frac{d v}{d t}

$v$ is the velocity and $a$ is the acceleration. Each is the rate of change of the one before.

Worked example

1
$s = 4 t^{3} - 6 t^{2} + 5 t$ — The displacement. We want $v$ and $a$ when $t = 2$ .
2
$v = \frac{d s}{d t} = 12 t^{2} - 12 t + 5$ — Differentiate $s$ : $4 t^{3} \to 12 t^{2}$ , $- 6 t^{2} \to - 12 t$ and $5 t \to 5$ .
3
$a = \frac{d v}{d t} = 24 t - 12$ — Differentiate $v$ : $12 t^{2} \to 24 t$ , $- 12 t \to - 12$ and the 5 becomes 0.
4
$v = 12 (2)^{2} - 12 (2) + 5$ — Put $t = 2$ into the velocity formula.
5
$v = 48 - 24 + 5 = 29 m/s$ — 12 × 4 = 48 and 12 × 2 = 24.
6
$a = 24 (2) - 12 = 36 m/s^{2}$ — Put $t = 2$ into the acceleration formula: 48 − 12 = 36.

Key idea

“At rest” means v = 0. “Constant speed” means a = 0. A negative velocity means moving the other way.

Which Method? Read the Question

Almost every question starts with $\frac{d y}{d x}$ .

You see	You do
“Find dy/dx”	Power rule on each term
A fraction or brackets	Rewrite as powers first
“Gradient is 8”	Set dy/dx = 8, solve for x
A point and a gradient	One equation from each
A given dy/dx with letters in y	match the powers, then the numbers in front
s in terms of t	v = ds/dt, a = dv/dt
“Turning point”	See Stationary Points

Key idea

Check one term by eye: $5 x^{3}$ must become $15 x^{2}$ .

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