The document discusses differentiation and tangents. It explains that differentiation finds the gradient of a curve at a point and is needed because curves have changing gradients. It provides examples of differentiating simple functions like y=3x^5. Tangents are lines that touch a curve at a single point, with the same gradient as the curve at that point. To find the equation of a tangent, take the derivative and plug in the x-value to get the slope, then use the point to find the y-intercept. Normals are perpendicular to tangents, with a gradient equal to the negative reciprocal of the tangent's gradient.

1. Starter
The line L passes through the points (0, 7) and
(3, 19). Work out the equation of the line L.

2. Starter
The line L passes through the points (0, 7) and
(3, 19). Work out the equation of the line L.
Gradient =
19−7
3−0
=
12
3
= 4
Equation: y=4x+c
Passes through (0,7) so y-intercept is 7
Therefore the equation is y=4x+7
How would we have worked out c if we were not given the y-intercept?

3. Calculus - Differentiation
• Differentiation is a way of finding a gradient at
a point on a curve.
• It is needed as curves have (by definition) a
constantly changing gradient.

4. Differentiation
• Why and how differentiation works is not
required knowledge for the Further Maths
exam
• It will not be covered in this session (look up
Differentiation from First Principles if you
want some light summer reading)

5. Differentiation – How To…
OR:
Multiply the whole thing by the power and
reduce it by one
𝑦 = 𝑎𝑥 𝑛
𝑑𝑦
𝑑𝑥
= a𝑥 𝑛−1

6. Differentiation - Example
𝑦 = 3𝑥5
𝑑𝑦
𝑑𝑥
= 5 × 3𝑥5−1
= 15𝑥4

7. Differentiation – Try these

8. Differentiation – Try these
𝟓𝒙 𝟒
𝟏𝟐𝒙 𝟑
𝟖𝒙 𝟑
+6x

9. Differentiation - Tangents
A tangent is a line that
touches a curve at a single
point.
The gradient of the
tangent is equal to the
gradient of the curve at
that point.

10. Differentiation - Tangents
As a straight line the
equation of the tangent is:
y=mx + c
This is equal to evaluated at the point P.
𝒅𝒚
𝒅𝒙
You will need to know a specific point on the
line to find c.

11. Equations of Tangents
Find the equation of the tangent to
𝑦 = 𝑥3
− 2𝑥2
+ 3𝑥, when x=2

12. Equations of Tangents
Find the equation of the tangent to
𝑦 = 𝑥3
− 2𝑥2
+ 3𝑥, when x=2
𝒅𝒚
𝒅𝒙
= 𝟑𝒙 𝟐
− 𝟒𝒙 + 𝟑 when x=2
𝒅𝒚
𝒅𝒙
=
𝟑(𝟐) 𝟐
−𝟒 𝟐 + 𝟑 = 𝟕 => y=7x+c
Then when x=2, 𝒚 = 𝟐 𝟑
− 𝟐 𝟐 𝟐
+ 𝟑 = 𝟑
So the line passes through (2 , 3)

13. Equations of Tangents
y=7x+c passing through (2,3)
So
3=7(2)+c
c=-11
Equation is y=7x-11

14. Equations of Tangents
Complete the table (only the tangents side)

15. Tangents and Normals
• A normal is a line that is perpendicular to the
tangent at a specific point.
• The gradient of a normal is the negative
reciprocal of the tangent (-1/tangent gradient)

16. Tangents and Normals
Tangent Gradient Normal Gradient
4 -1/4
-3 1/3
1/2 -2
-1/3 3
3/4 -4/3
-7/2 2/7
Examples of negative
reciprocals
Once you have worked
out the gradient, finding
the equation is exactly the
same as for the tangent

17. Tangents and Normals
Now complete the Normals side of the sheet