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Chapter : Differentiation
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)
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)
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)
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)
Question 1
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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)
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Question 3
If y = 2s⁶ and x = 2s -1, find dy/dx in
so we replace with S,
terms of s.
=6
From
x = 2s – 1
2s = x + 1
S =x+1
= 6s⁵
2
Therefore y = 2s⁶
y=2
dy/dx = 12
1
X 2
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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)
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Question 5
Find the coordinates of the point on
the curve y = 32x² + 1/x at which
the tangent is parallel to the xaxis.
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)
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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)
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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
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)
y = x² + 3x
dy/dx = 2x + 3
δx = 2.01 – 2
= 0.01
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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)
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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)
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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.
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
0.8cms¯¹
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