1. Chapter : Differentiation

2. Questions 1. Find the value of Answers (1) 2. If f(x) = 2[2x + 5]⁴, find f ‘ (-2) Answers(2) 3. If y = 2s⁶ and x = 2s – 1, find dy/dx in terms of s Answers(3) 4. Given that y = 15x [3-x], calculate a) the value of x when y is maximum b) the maximum value of y Answers(4)

3. Questions 5. Find the coordinate of the point on the curve y = 32x² +1 at which the tangent is parallel with the x-axis Answers(5) 6. Find the coordinates of the points on the curve y = [3x – 1]³ where the gradient of the tangent to the curve is 81 Answers (6) 7. Given that y = x + 3x, use differentiation to find the small change in y when x increases 2 to 2.01. Answers (7)

4. Questions 8. Find the equation of the normal to the curve y = 2x² - 3x+2 at the point where its x coordinate is 3. Answers(8) 9. Find the coordinates of the turning point of y = x³ + 15 x² - 24, 2 and state whether it is a maximum or minimum point. Answers (9)

5. Questions 10. The radius of a ball is increasing at a rate of 0.8cm¯¹. Find the rate of change of its surface area when its radius is 8cm. Answer (10)

6. Question 1 Back To Questions

7. Question 2 If f(x) = 2[2x + 5]⁴, find f ’ (-2) If f(x) =2[2x + 5]⁴ f ’ (x) = 8[2x + 5]³. 2 = 16 [2x + 5]³ f ’ (-2) = 16[2(-2) + 5]³ = 16[1]³ = 16 answer (2) Back To Questions

8. Question 3 If y = 2s⁶ and x = 2s -1, find dy/dx in terms of s. From x = 2s – 1 2s = x + 1 S =x+1 2 Therefore y = 2s⁶ y = 2 dy/dx = 12 1 X 2 so we replace with S, = 6 = 6s⁵ Back to Questions

9. Question 4 Given that y=15x [3 – x] calculate a) The value of x when y is a maximum From y = 15x [3-x] = 45x – 15x² dy/dx = 45-30x = 0 45 = 30x x = 3/2 answer (a) b) The maximum value of y Substitute x = 3/2 into the equation y = 15x [3 –x] = 15(3/2) (3-3/2) = 45/2 [3/2] = 33.75 answer (b) Back to Questions

10. Question 5 Find the coordinates of the point on the curve y = 32x² + 1/x at which the tangent is parallel to the x- axis. y = 32x²+ 1/x = 32x²+ x¯¹ dy/dx = 64x –x¯² = 64x - 1/x² = 0 64x = 1/x² 64x³=1 x³=1/64 x = ∛(1/64) = 1/4 when x = ¼ y = 32x²+ 1/x substitute x = ¼ into x in the equation y = 32( 1/4 )²+ 1/(1/4) y = 32 ( 1/16) + 4 y = 32/16 + 4 = 96/16 = 6 therefore, the coordinates of the point is (x, y ) = (1/4, 6) answer (5) Back to Questions

11. Question 6 Find the coordinates of the points on the curve y = [3x-1]³ where the gradient of the tangent to the curve is 81 y = [3x-1]³ dy/dx = 3[3x-1] ². 3 = 9[3x-1]² given dy/dx = 81 9[3x-1]²=81 [3x-1]=9 3x-1=√9 3x-1=3 3x=4 x=4/3 with x = 4/3 y = [3(4/3)-1]³ = 3³ = 27 so, the coordinates of the point is [4/3, 27] answer (6) Back to Questions

12. Question 7 Given that y = x+3x, use differentiation to find the small change in y when x increases from 2 to 2.01. δy/δx = dy/dx δy = dy/dx × δx y = x² + 3x dy/dx = 2x + 3 δx = 2.01 – 2 = 0.01 substitute the information into the formula δy=[2x+3]×0.01 = 2[2]+3×0.01 (original x is 2) = 7×0.01 = 0.07 answer (7) Back to Questions

13. Question 8 Find the equation of the normal to the curve y = 2x² - 3x + 2 at the point where its x coordinate is 3. y = 2x² - 3x + 2 dy/dx = 4x + 3 substitute x = 3 into y y = 2[3]² + 3[3] + 2 = 11 coordinate of the point is (3, 11) substitute x = 3 into dy/dx dy/dx = 4[3]- 3 = 9 So, the gradient of the curve at the point is 9 Thus, the gradient of the normal to that is -1/9 General form of equation y - y1 = m[x-x1] y – 11 = -1/9 [x-3] gradient with negative regression y= -1/9 x + 34/3 y- intersect answer (8) Back to Questions

14. Question 9 Find the coordinates of the turning point of y = x³ +15/2 x² - 24, and state whether it is a maximum or minimum point. From y = x³ +15/2 x² - 24 dy/dx = 3x²+15x At turning point, dy/dx = 0 3x² + 15x = 0 x[3x+15] = 0 so, x=0 or 3x+15 = 0 3x = -15 x = -5 Coordinates of turning point when y = [0]³ + 15/2 [0]² - 24 = -24 Coordinates of the point (0, -24) Coordinates of turning point when x = -5 y = [-5]³ + 15/2 [-5]² - 24 = -24 y = -125 + 375/2 – 24 = 77/2 Coordinates of the point (-5, 77/2) Substitute the coordinates to determine the min and max points At point (0,-24) dy/dx = 3x² + 15x dy²/dx² = 6x + 15 = 6[0] + 15 = 15>0 (minimum point) At point (-5, 77/2) dy/dx = 3x² + 15x dy²/dx² = 6x + 15 = 6[-5] + 15 = -15 < 0 (maximum point) Back to Questions

15. Question 10 The radius of a ball is increasing at a rate of 0.8cms¯¹. Find the rate of change of its surface area when its radius is 8cm. 0.8cms¯¹ As we know the area of sphere = 4πr² Rate of change its surface area (dA/dt = dA/dr × dr/dt) A = 4πr² dA/dr = 8πr dr/dt = 0.8cms¯¹ r = 8 dA/dt = 8πr × 0.8 = 8π[8] × 0.8 = 64π × 0.8 = 51.2πcm²s¯¹ Back to Questions