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Differentiation

Differentiation

Using differentiation: gradients and tangents

Tangents And Gradients Extended 8 min read

On this page
  1. In this note
  2. Formula bank
  3. Gradient at a point
  4. Where is the gradient equal to a value?
  5. The equation of a tangent
  6. Ways they test it
  7. Mistakes to avoid
  8. Summary
  9. Practice
  10. Can I do this?
  11. Quiz

In this note

Syllabus E2.12: using differentiation for gradients
  • E2.12Find the gradient of a curve at a given point.
  • E2.12Find the point(s) where the gradient has a given value.
  • E2.12Find the equation of the tangent at a point.

Before you start: the gradient function and the equation of a straight line. Next: turning points.

Formula bank

Gradient at a pointLearn itput \(x=a\) into \(\dfrac{dy}{dx}\)the answer is the gradient of the curve there.Use it when "find the gradient at x = ...".
Point with a given gradientLearn itsolve \(\dfrac{dy}{dx}=k\)then y from the CURVE's equation.Use it when "find the point where the gradient is ...".
Straight lineLearn it\(y=mx+c\)\(m\) = gradient, \(c\) = where it crosses the y-axis.Use it when you need the equation of a tangent.

Gradient at a point

General methodHow to find the gradient at a point

  1. Differentiate to get \(\dfrac{dy}{dx}\).
  2. Put the x-value into \(\dfrac{dy}{dx}\).

Example Find the gradient of \(y=2x^{3}-x\) at \(x=2\).

Differentiate, then put x = 2 into dy/dx.

  1. \(\dfrac{dy}{dx}=6x^{2}-1\)
  2. \(6(4)-1=23\)

Answer 23

Try it yourself

1
Gradient of \(y=x^{2}-5x+2\) at \(x=2\).
Hint
\(\tfrac{dy}{dx}=2x-5\).
Show answer

Answer −1

Where is the gradient equal to a value?

General methodHow to find the point where the gradient is k

  1. Differentiate.
  2. Put \(\dfrac{dy}{dx}=k\) and solve for \(x\) (it may be a quadratic with two answers).
  3. Find \(y\) by putting each \(x\) into the ORIGINAL equation of the curve.
  4. Write each point as \((x,\ y)\).

Example Find the coordinates of the point on \(y=x^{2}-4x+1\) where the gradient is 2.

Set dy/dx = 2, solve, then y from the curve.

  1. \(\dfrac{dy}{dx}=2x-4\)
  2. \(2x-4=2\Rightarrow x=3\)
  3. \(y=9-12+1=-2\)

Answer \((3,\ -2)\)

Try it yourself

1
Where is the gradient of \(y=x^{2}+6x\) zero? \(x=\)
Show answer

Answer \(x=-3\)

2
Find the point on \(y=x^{2}-2x\) where the gradient is 4.
Hint
\(2x-2=4\).
Show answer

Answer \((3,\ 3)\)

The equation of a tangent

A tangent is the straight line that just touches the curve at a point, with the same gradient as the curve there. So you need a point and a gradient, then \(y=mx+c\).

General methodHow to find the equation of the tangent at x = a

  1. Point: put \(x=a\) into the curve to get \(y\).
  2. Gradient: put \(x=a\) into \(\dfrac{dy}{dx}\) to get \(m\).
  3. Put \(x\), \(y\) and \(m\) into \(y=mx+c\) and solve for \(c\).
  4. Write \(y=mx+c\) with the numbers.
tangent at x = 3-224-4-2246810xy(3, 4)
Tangent \(y=9x-23\) at (3, 4).

Example Find the equation of the tangent to \(y=x^{3}-3x^{2}+4\) at \(x=3\).

Point, gradient, then y = mx + c.

  1. point: \(y=27-27+4=4\), so \((3,\ 4)\)
  2. \(\dfrac{dy}{dx}=3x^{2}-6x\); at 3: \(27-18=9\)
  3. \(4=9(3)+c\Rightarrow c=-23\)

Answer \(y=9x-23\)

Try it yourself

1
Tangent to \(y=x^{2}+1\) at \(x=2\): \(y=mx+c\). Find \(m\) and \(c\).
Hint
Point (2, 5); gradient \(2x=4\).
Show answer

Answer \(y=4x-3\)

Ways they test it

Type 1 Gradient at a point

"Find the gradient of the curve \(y=\dots\) at the point where \(x=2\)." (2-3 marks)

Spot it "gradient" with an x-value.

  1. Differentiate.
  2. Substitute the x-value into dy/dx.

Type 2 Coordinates where the gradient is k

"Find the coordinates of the points where the gradient is 9." (4 marks)

Spot it "coordinates" and a gradient value.

  1. Solve dy/dx = k.
  2. y from the curve: the last mark is for the y-values.

For example \(y=x^{3}-3x^{2}+4\), gradient 9

  1. \(3x^{2}-6x=9\Rightarrow x^{2}-2x-3=0\Rightarrow x=3,\ -1\)
  2. \(y(3)=4\), \(y(-1)=0\)

Answer \((3,\ 4)\) and \((-1,\ 0)\)

Type 3 Equation of the tangent

"Find the equation of the tangent to the curve at \(x=3\)." (4 marks)

Spot it "tangent" and a point.

  1. Point from the curve.
  2. m from dy/dx.
  3. y = mx + c.

Mistakes to avoid

  • ✗ Finding y by putting x into dy/dx✓ put x into the curve's equationdy/dx gives the gradient, not the height.
  • ✗ Using the y-value as the gradient of the tangent✓ \(m\) is \(\dfrac{dy}{dx}\) at that pointHeight and steepness are different things.
  • ✗ Stopping at x = 3, −1 when asked for coordinates✓ \((3,\ 4)\) and \((-1,\ 0)\)Coordinates need both x and y.

Summary

  • Gradient at a point: substitute its x into \(\dfrac{dy}{dx}\).
  • Gradient equal to \(k\): solve \(\dfrac{dy}{dx}=k\), then y from the curve.
  • Tangent: point (from the curve) + gradient (from \(\dfrac{dy}{dx}\)) → \(y=mx+c\).

If you see …

  • "gradient at"differentiate, substitute.
  • "where the gradient is"set dy/dx equal to it and solve.
  • "tangent"point + gradient → y = mx + c.

Practice

Practice: easy, medium, hard

Easy

1
Find the gradient of \(y=x^{3}\) at \(x=-2\).
Hint
\(3x^{2}\): a square is positive.
Show working
  1. \(3x^{2}\)
  2. \(3(4)=12\)

Answer 12

2
Find the gradient of \(y=5x^{2}-3x\) at \(x=1\).
Show working
  1. \(10x-3\)
  2. \(10-3=7\)

Answer 7

3
Where is the gradient of \(y=x^{2}+6x\) equal to 0? \(x=\)
Show working
  1. \(2x+6=0\)
  2. \(x=-3\)

Answer \(-3\)

Medium

1
Find the coordinates of the point on \(y=x^{2}+3x\) where the gradient is 7.
Hint
\(2x+3=7\).
Guide me

What is it asking? a point (x and y)

What do we know? the gradient is 7

Plan solve dy/dx = 7, then y from the curve

Show working
  1. \(2x+3=7\Rightarrow x=2\)
  2. \(y=4+6=10\)

Answer \((2,\ 10)\)

2
Find the equation of the tangent to \(y=2x^{2}-x+3\) at \(x=1\).
Hint
Point (1, 4).
Guide me

What is it asking? the tangent y = mx + c

What do we know? x = 1

Plan point from the curve, m from dy/dx, then c

Show working
  1. point: \(2-1+3=4\)
  2. \(\tfrac{dy}{dx}=4x-1\); at 1: \(m=3\)
  3. \(4=3+c\Rightarrow c=1\)

Answer \(y=3x+1\)

3
Find the gradient of \(y=(x+2)(x-5)\) at \(x=4\).
Hint
Expand first.
Guide me

What is it asking? the gradient at x = 4

What do we know? brackets

Plan expand, differentiate, substitute

Show working
  1. \(y=x^{2}-3x-10\)
  2. \(\tfrac{dy}{dx}=2x-3\)
  3. \(8-3=5\)

Answer 5

Hard

1
Find the coordinates of the two points on \(y=x^{3}-12x\) where the gradient is 15.
Hint
\(3x^{2}-12=15\Rightarrow x^{2}=9\).
Guide me

What is it asking? two points

What do we know? gradient 15

Plan solve dy/dx = 15 (two x-values), y from the curve for each

Show working
  1. \(3x^{2}-12=15\)
  2. \(x^{2}=9\Rightarrow x=3\) or \(-3\)
  3. \(y(3)=27-36=-9\); \(y(-3)=-27+36=9\)

Answer \((3,\ -9)\) and \((-3,\ 9)\)

PartAnswer / workingMarksGuidance
\(3x^{2}-12\)B1
\(3x^{2}-12=15\)M1
\(x=\pm3\)A1
\((3,\ -9)\), \((-3,\ 9)\)A1
Total: 4 marks
2
Find the equation of the tangent to \(y=x^{3}-2x^{2}+x\) at \(x=2\).
Guide me

What is it asking? the tangent

What do we know? x = 2

Plan point, gradient, y = mx + c

Show working
  1. point: \(8-8+2=2\)
  2. \(\tfrac{dy}{dx}=3x^{2}-4x+1\); at 2: \(12-8+1=5\)
  3. \(2=10+c\Rightarrow c=-8\)

Answer \(y=5x-8\)

PartAnswer / workingMarksGuidance
\(3x^{2}-4x+1\)B1
\(m=5\) and \((2,\ 2)\)M1
\(y=5x-8\)A1
Total: 3 marks
3
The gradient of \(y=x^{2}-2x\) at the point P is 4. Find the coordinates of P.
Guide me

What is it asking? the point P

What do we know? gradient 4 at P

Plan solve 2x − 2 = 4, then y

Show working
  1. \(2x-2=4\Rightarrow x=3\)
  2. \(y=9-6=3\)

Answer \((3,\ 3)\)

PartAnswer / workingMarksGuidance
\(2x-2=4\)M1
\(x=3\)A1
\((3,\ 3)\)A1
Total: 3 marks

Can I do this?

Anything unticked: revise that section first. All ticked: take the quiz.

Quiz

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