1. Derivera Polynom We have looked many times at the graph f(x)=x 2. A tangent is a line which touches a curve at one point only. In the graph f(x) = x 2 you can see a tangent at the point (1;1). You could draw a tangent at (2;4) to find the gradient where x=2. Note that f(2) = x 2 = 4 The gradient at any point on f(x) is given by the gradient (lutning) of the tangent to that point.

2. Derivera Polynom The gradient (lutning) at any point on f(x) is given by the gradient (lutning) of the tangent to that point. We do not need to draw the tangent ourselves if we know f(x) in order to find the gradient (lutning).

3. Derivera Polynom If we draw tangents at (1;1), (2;4); (3;9) and (4;16) trust me when I tell you that you will discover the results below! xf(x)f’(x) 112 244 396 4168

4. Derivera Polynom Can you see a pattern between x and f’(x) for f(x)=x 2 ? xf(x)f’(x) 112 244 396 4168

5. Derivera Polynom If f(x) = x 2 then f’(x) = 2x. NOTE: WE will PROVE this later in the term as the syllabus says we must do. If f(x) = x 3 then f’(x) = 3x 2 If f(x) = x 4 – 2x 2 then f’(x) = 4x 3 – 4x Can you see the pattern yet? (Remember that 4x=4x 1)

6. Derivera Polynom Here is the rule! If f(x) = ax n  f’(x) = nax n-1 Example f(x) = 3x 6 a=3, n=6 so f’(x) = 3.6x 6-1  f’(x) = 18x 5 Think of the rule as 'multiply by the index, then reduce the index by 1'.

7. Try These For the following functions f(x), find f’(x) a)f(x) = x 2 +7x b)f(x) = r 2 c)f(x) = 16x 3 – 4x 2 + 2x d) f(x) = 3x(4x + 2) Think of the rule as 'multiply by the index, then reduce the index by 1'.

8. Answers a)f’(x) = 2x+7 b)f’(x) = 2 r c)f’(x) = 48x 2 – 8x + 2 d)f’(x) = 24x + 6 Think of the rule as 'multiply by the index, then reduce the index by 1'.

9. Other Important Facts What is f’(x) if f(x) = x 3 + x 2 – 10? Write f(x) as x 3 + x 2 – 10x 0 since 10 0 = 1  f’(x) = 3x 2 + 2x NOTE: f’(k) = 0 where k is any number, for example, 25.