Chapter : Differentiation

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

Questions 5. Find the coordinate of the point on the curve y = 32x +1 at which the tangent is parallel with the x-²axis Answers(5)6. Find the coordinates of the points on the curve y = [3x – 1] where the gradient of the tangent to the ³curve 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)

Questions 8. Find the equation of the normal to the curve y = 2x - 3x+2 at the point where its ² x coordinate is 3.

Answers(8)9. Find the coordinates of the turning point of y = x + ³ 15 x - 24 , ² 2and state whether it is a maximum or minimum point. Answers (9)

Questions10. The radius of a ball is increasing at a rate of 0.8cm¯ . Find the rate of change of its surface ¹area when its radius is 8cm.

Answer (10)

Question 1

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Question 2If f(x) = 2[2x + 5] , find f ’ (-2)⁴

If f(x) =2[2x + 5] ⁴ f ’ (x) = 8[2x + 5] . 2³ = 16 [2x + 5] ³ f ’ (-2) = 16[2(-2) + 5]³ = 16[1]³ = 16 answer (2)

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Question 3If y = 2s and x = 2s -1, find ⁶dy/dx in terms of s.From x = 2s – 12s = x + 1 S =x+1 2Therefore y = 2s ⁶y = 2

dy/dx = 12 1 X 2

so we replace with S,= 6 = 6s⁵

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Question 4Given that y=15x [3 – x] calculate a) The value of x when y is a maximumFrom y = 15x [3-x] = 45x – 15x² dy/dx = 45-30x = 0 45 = 30x x = 3/2 answer (a)

b) The maximum value of ySubstitute x = 3/2 into the equation y = 15x [3 –x] = 15(3/2) (3-3/2) = 45/2 [3/2] = 33.75 answer (b)

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Question 5Find the coordinates of the point on the curve y = 32x ²+ 1/x at which the tangent is

parallel to the x-axis.

y = 32x + 1/x² = 32x + x¯² ¹ dy/dx = 64x –x¯ ² = 64x - 1/x = 0² 64x = 1/x ² 64x =1³ x =1/64³ x = (1/64)∛ = 1/4

when x = ¼ y = 32x + 1/x²

substitute x = into x in the ¼equation y = 32( 1/4 ) + 1/(1/4)² y = 32 ( 1/16) + 4 y = 32/16 + 4 = 96/16 = 6therefore, the coordinates of the point is (x , y ) = (1/4 , 6) answer (5)Back to

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Question 6Find the coordinates of the points on the curve y = [3x-1] where the gradient of ³the tangent to the curve is 81 y = [3x-1]³ dy/dx = 3[3x-1] . 3² = 9[3x-1]² given dy/dx = 81 9[3x-1] =81² [3x-1]=9 3x-1=√9 3x-1=3 3x=4 x=4/3

with x = 4/3 y = [3(4/3)-1]³ = 3 = 27³so, the coordinates of the point is [4/3 , 27] answer (6)Back to Questions

Question 7Given 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 = x + 3x² 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)

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Question 8 Find the equation of the normal to the curve y = 2x - 3x + 2 at the ²point where its x coordinate is 3. y = 2x² - 3x + 2 dy/dx = 4x + 3 substitute x = 3 into y y = 2[3] + 3[3] + 2² = 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 9Thus, the gradient of the normal to that is -1/9General 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)

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Question 9Find the coordinates of the turning point of y = x +15/2 ³x - 24, and state whether it is ²a maximum or minimum point.From y = x +15/2 x - 24³ ² dy/dx = 3x +15x² At turning point, dy/dx = 0 3x + 15x = 0² x[3x+15] = 0so, x=0 or 3x+15 = 0 3x = -15x = -5Coordinates of turning point when y = [0] + 15/2 [0] - ³ ²24 = -24Coordinates of the point (0, -24)

Coordinates of turning point when x = -5 y = [-5] + 15/2 [-5] - 24 = -³ ²24 y = -125 + 375/2 – 24 = 77/2Coordinates of the point (-5, 77/2)Substitute the coordinates to determine the min and max pointsAt point (0,-24)dy/dx = 3x + 15x²dy /dx = 6x + 15 ² ² = 6[0] + 15 = 15>0 (minimum point)At point (-5, 77/2)dy/dx = 3x + 15x²dy /dx = 6x + 15 ² ²= 6[-5] + 15= -15 < 0 (maximum point) Back to

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Question 10The radius of a ball is increasing at a rate of 0.8cms¯ . Find the ¹rate of change of its surface area when its radius is 8cm.

0.8cms¯¹

As we know the area of sphere = 4πr²Rate 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 = 51.2πcm s¯² ¹

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