Differentiation
Differentiation: the gradient function
In this note
- E2.12Know that \(\dfrac{dy}{dx}\) gives the gradient of a curve at any point.
- E2.12Differentiate \(ax^{n}\) (\(n\) a positive whole number or 0) and sums of up to three such terms.
- E2.12Find the gradient of a curve at a given point.
Before you start: expanding brackets and gradients of curves. Next: gradients and tangents.
Formula bank
What dy/dx means
A curve is steeper in some places than others. Draw tangents to \(y=x^{2}\) at \(x=1, 2, 3\) and their gradients are \(2, 4, 6\): always \(2\times x\).
That rule, gradient \(=2x\), is called the derivative. It is written \(\dfrac{dy}{dx}\) (say "dee y by dee x"). Differentiating means finding it.
Differentiating one term
General methodHow to differentiate any term \(ax^{n}\)
- Multiply the number in front by the power: \(4x^{3}\to3\times4=12\).
- Take 1 off the power: \(x^{3}\to x^{2}\). So \(4x^{3}\to12x^{2}\).
- A term in \(x\) (like \(7x\)) just leaves its number: \(7\).
- A number on its own (like \(9\)) disappears: it becomes \(0\).
| \(y\) | \(\dfrac{dy}{dx}\) | How |
|---|---|---|
| \(x^{2}\) | \(2x\) | \(2\times x^{1}\) |
| \(x^{5}\) | \(5x^{4}\) | power in front, then one less |
| \(4x^{3}\) | \(12x^{2}\) | \(3\times4=12\) |
| \(-3x^{2}\) | \(-6x\) | \(2\times(-3)=-6\) |
| \(\tfrac12x^{4}\) | \(2x^{3}\) | \(4\times\tfrac12=2\) |
| \(7x\) | \(7\) | \(x\) goes |
| \(9\) | \(0\) | a number goes |
Try it yourself
Show answer
Answer \(5x^{4}\)
Hint
Show answer
Answer \(12x^{2}\)
Show answer
Answer \(-6x\)
Several terms, and brackets
General methodHow to find \(\dfrac{dy}{dx}\) for any expression
- Brackets? Expand them first, so every term is a number times a power of x.
- Differentiate each term on its own with the rule. Keep the + and − signs.
- Write the answer as one expression.
Example Find \(\dfrac{dy}{dx}\) when \(y=2x^{3}-5x^{2}+3x-8\).
Differentiate term by term.
- \(2x^{3}\to6x^{2}\)
- \(-5x^{2}\to-10x\)
- \(3x\to3\)
- \(-8\to0\)
Answer \(\dfrac{dy}{dx}=6x^{2}-10x+3\)
Example Find \(\dfrac{dy}{dx}\) when \(y=(x+3)(x-2)\).
Expand first, then differentiate each term.
- \(y=x^{2}+x-6\)
Answer \(\dfrac{dy}{dx}=2x+1\)
Try it yourself
Show answer
Answer \(3x^{2}+2\)
Show answer
Answer \(10x-4\)
Hint
Show answer
Answer \(2x+4\)
The gradient at a point
General methodHow to find the gradient of a curve at a point
- Find \(\dfrac{dy}{dx}\).
- Put the \(x\)-value of the point into \(\dfrac{dy}{dx}\) (NOT into \(y\)).
- Positive = the curve goes up there; negative = it goes down; 0 = it is flat.
Example Find the gradient of \(y=x^{4}-6x^{2}+2\) at \(x=-1\).
Differentiate, then put x = −1 into dy/dx.
- \(\dfrac{dy}{dx}=4x^{3}-12x\)
- \(4(-1)^{3}-12(-1)=-4+12=8\)
Answer 8
Try it yourself
Hint
Show answer
Answer 8
Show answer
Answer −12
Ways they test it
Type 1 Find dy/dx
"\(y=3x^{4}-2x^{2}+7\). Find \(\dfrac{dy}{dx}\)." (2 marks)
Spot it the words "find dy/dx" or "differentiate".
- Differentiate each term.
- A number on its own goes; one mark per correct term.
For example \(y=3x^{4}-2x^{2}+7\)
- \(12x^{3}-4x\)
Answer \(12x^{3}-4x\)
Type 2 Expand, then differentiate
"Differentiate \((2x-1)^{2}\)."
Spot it brackets in the expression.
- Expand.
- Differentiate each term.
For example \((2x-1)^{2}\)
- \(4x^{2}-4x+1\)
Answer \(8x-4\)
Type 3 Gradient at a point
"Find the gradient of the curve at \(x=2\)."
Spot it "gradient" and an x-value.
- Find dy/dx.
- Substitute the x-value.
Type 4 Fractions in front
"\(y=\tfrac13x^{3}+x^{2}-5\). Find \(\dfrac{dy}{dx}\)."
Spot it fractions as the numbers in front.
- Multiply the fraction by the power: \(3\times\tfrac13=1\).
- Take 1 off the power.
For example \(y=\tfrac13x^{3}+x^{2}-5\)
- \(x^{2}+2x\)
Answer \(x^{2}+2x\)
Mistakes to avoid
- ✗ \(9\to9\)✓ \(9\to0\)A number on its own has gradient 0.
- ✗ \(4x^{3}\to4x^{2}\)✓ \(12x^{2}\)Multiply by the power as well as taking 1 off.
- ✗ \(7x\to7x\) or \(7x\to0\)✓ \(7x\to7\)\(x=x^{1}\): \(1\times7\), and \(x^{0}=1\).
- ✗ \((x+3)(x-2)\to(1)(1)=1\)✓ expand first: \(2x+1\)The rule works on single powers of x, not on brackets.
- ✗ Putting the x-value into y for the gradient✓ put it into \(\dfrac{dy}{dx}\)y gives the height, dy/dx gives the gradient.
Summary
- \(\dfrac{dy}{dx}\) is the gradient of the curve at any point.
- Each term: multiply by the power, then take 1 off the power.
- \(kx\to k\); a number on its own \(\to0\).
- Expand brackets first.
- Gradient at a point: put its x-value into \(\dfrac{dy}{dx}\).
If you see …
- "find dy/dx" or "differentiate"term by term with the rule.
- bracketsexpand first.
- "gradient at x = ..."substitute into dy/dx.
Practice
Practice: easy, medium, hard
Easy
Hint
Show working
- \(x^{2}\to2x\)
- \(3x\to3\)
Answer \(2x+3\)
Hint
Show working
- \(10x-4+0\)
Answer \(10x-4\)
Hint
Show working
- \(12x^{2}-1\)
- \(12-1=11\)
Answer 11
Medium
Hint
Guide me
What is it asking? dy/dx
What do we know? two terms, one with a fraction in front
Plan multiply each by its power, take 1 off
Show working
- \(\tfrac12x^{4}\to2x^{3}\)
- \(-3x\to-3\)
Answer \(2x^{3}-3\)
Guide me
What is it asking? dy/dx
What do we know? three terms; the 7 is a number on its own
Plan term by term
Show working
- \(12x^{3}-4x+0\)
Answer \(12x^{3}-4x\)
Hint
Guide me
What is it asking? the gradient at x = 3
What do we know? a bracket squared
Plan expand, differentiate, substitute
Show working
- \(y=4x^{2}-4x+1\)
- \(\tfrac{dy}{dx}=8x-4\)
- \(8(3)-4=20\)
Answer 20
Hard
Hint
Guide me
What is it asking? the gradient at x = −3
What do we know? a fraction in front of the first term
Plan differentiate, then substitute carefully with the negative
Show working
- \(\tfrac{dy}{dx}=x^{2}+2x\)
- \((-3)^{2}+2(-3)=9-6=3\)
Answer 3
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(x^{2}+2x\) | B2 | B1 for each term | |
| 3 | B1FT | ||
| Total: 3 marks | |||
Guide me
What is it asking? the gradient at x = 2
What do we know? four terms
Plan differentiate term by term, then substitute
Show working
- \(6x^{2}-10x+3\)
- \(24-20+3=7\)
Answer 7
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(6x^{2}-10x+3\) | B2 | B1 for two correct terms | |
| 7 | B1FT | ||
| Total: 3 marks | |||
Hint
Guide me
What is it asking? dy/dx and where it is 0
What do we know? a bracket times x
Plan expand, differentiate, solve 2x + 4 = 0
Show working
- \(y=x^{2}+4x\)
- \(\tfrac{dy}{dx}=2x+4\)
- \(2x+4=0\Rightarrow x=-2\)
Answer \(2x+4\); \(x=-2\)
| Part | Answer / working | Marks | Guidance |
|---|---|---|---|
| \(2x+4\) | B1 | ||
| \(2x+4=0\) | M1 | ||
| \(x=-2\) | A1 | ||
| Total: 3 marks | |||
Can I do this?
- I know dy/dx is the gradient of the curve.
- I can differentiate one term, a term in x, and a number on its own.
- I can differentiate up to three terms, expanding brackets first.
- I can find the gradient at a given point.
Anything unticked: revise that section first. All ticked: take the quiz.
Quiz
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