University:
University of Cambridge
Course:
0580 IGCSE | Mathematics
Academic year:

2024
Views:

167

Pages:

3
Author:

Shahbaz A.

6663/01 Edexcel GCE Core Mathematics C1 Advanced Subsidiary shahbaz ahmed July 2024 Differentiation: Tangents and Normals C1 Differentiation – Tangents and Normals 1. The curve C has equation y= (a) Find dy dx (x + 3)(x − 8) ,x > 0 x (1) in its simplest form. (b) Find an equation of the tangent to C at the point where x = 2 Solution 1 (a) y = (x+3)(x−8) x Let f (x) = (x + 3)(x − 8) d(x+3)(x−8) dx d(f (x)) dx = d(f (x)) dx d(x+3) = (x + 3) d(x−8) dx + (x − 8) dx d(f (x)) dx = (x + 3) + (x − 8) = 2x − 5 g(x) = x d(g(x)) dx = dx dx =1 Putting in the formula dy dx = (x)) g(x) d(fdx −f (x) d(g(x)) dx [g(x)]2 2 dy dx = x(2x−5)−(x+3)(x−8) x2 dy dx = 2x2 −5x−(x2 −8x+3x−24) x2 dy dx = 2x2 −5x−(x2 −5x−24) x2 dy dx = 2x2 −5x−x2 +5x+24 x2 dy dx = 2x2 −x2 +24 x2 dy dx = x2 +24 x2 dy (b) m= dx = x2 +24 x2 Where m is the slope of tangent at any point x. Putting x=2 dy = m= dx 22 +24 22 =7 Also Putting x=2 in the equation y= (2+3)(2−8) 2 = − 30 2 = −15 Equation of the line passing through (2, −15) with slope m = 7 y − (−15) = 7(x − 2) y + 15 = 7x − 14 7x − 14 = y + 15 7x − 14 − y − 15 = 0 7x − y − 29 = 0 3