1. Method for finding tangent lines
Pierre de Fermat ( )
Method for finding tangent lines

2. Life and work Interesting facts: Contribution to calculus development:
He was a French lawyer
He pursued maths as a hobby
He is known as the “Prince of Amateurs”
His contribution to calculus was less well known.
Contribution to calculus development:
determining maxima, minima
finding tangents to various curves
Evaluating area under a graph

3. Fermat’s Similar Triangles
Tangent line at point T

4. Fermat’s Similar Triangles
T(x,y)
A(x,0) s y
O(x-s,0)
Fermat used 𝑇𝐴 𝑂𝐴 to calculate the gradient at T
𝑇𝐴 𝑂𝐴 = 𝑦−0 𝑥−(𝑥−𝑠) = 𝑓(𝑥) 𝑠

5. Fermat’s Similar Triangles𝑇
To calculate value of s, he used a technique based on similar triangles.

6. Fermat’s Similar TrianglesO A
He starts drawing tangent line at point T
Then create triangle ∆OAT.

7. Fermat’s Similar TrianglesP O B A
Then extend the length E to create a new triangle ∆ OPB

8. Fermat’s Similar TrianglesP O B A s E
Set the length OA =s, OB = s + E

9. Fermat’s Similar TrianglesO A T P B s E
Properties of similar triangle:
Perimeters of similar triangles are in the same ratio as their corresponding sides 
𝑠 𝑠+𝐸 = 𝑇𝐴 𝑃𝐵

10. Fermat’s Similar TrianglesO A T P B s
𝐸→0 O A T P B s E
Assume T has a coordinate (x,y) where y = f(x)
At this point, Fermat claims that error disappears when 𝐸→0
When 𝐸→0, PB →𝑓 𝑥 + 𝐸
𝑠 𝑠+𝐸 = 𝑇𝐴 𝑃𝐵 = 𝑓(𝑥) 𝑓 𝑥 + 𝐸

11. Fermat’s Similar TrianglesO A T P B s
𝐸→0
By rearranging the equation , 
s= 𝑓(𝑥) 𝑓 𝑥 + 𝐸 −𝑓 𝑥 /𝐸

12. Fermat’s Similar TrianglesO A T P B s
𝐸→0
So, the gradient of the tangent at point T can be found as follows:- 
𝑓(𝑥) 𝑠

13. Example
Let’s find the tangent of the curve at (2,4)

14. Example Let find the tangent of the curve y= 𝑥 2 at (2,4)
s= 𝑥 2 ((𝑥+𝐸) 2 − 𝑥 2 )/𝐸 = 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸

15. s= 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸 = 𝑥 2 𝐸+2𝑥 When 𝐸→0, then S = 𝑥 2
Example
s= 𝑥 2 ( 𝑥 2 +𝐸 2 +2𝑥𝐸− 𝑥 2 )/𝐸 = 𝑥 2 𝐸+2𝑥 When 𝐸→0, then S = 𝑥 2

16. Example
Gradient of the tangent = 𝑓(𝑥) 𝑠 = 𝑥 2 𝑥/2 = 2𝑥 At point (2,4), Gradient of the tangent = 2 x 2 = 4

17. Example Let y = 4x+c be the equation of the tangent at point (2,4)
By substituting (2,4) in the equation,
4=4 (2) + C
C = -4

18. Example
Therefore, the equation of the tangent is
Y = 4 x -4