Name: xxxxxxxxxxxxxxx
ID: xxxxxxxxxxx
Class: xxxx
INDIVIDUAL ASSIGNMENT
1. The balance of a bank account after 3 years with respect to the interest rate r% is given by
A= 1000(1+r/1000)36 Find the rate of change of the balance with respect to r when r = 6
SOLUTION
A=1000(1+6/1000)
=1000(1+0.006)
36
36
= 1000(1.24)
=1240.3
36= 3 YEARS
12= 1 YEAR
A=1000(1+6/1000)
12
=1074.42
AVERAGE RATE OF CHANGE IS GIVEN BY: Y3-Y1/X3-X1
1240.3-1074.42/36-12
ARC = 6.912
2- Find f (a) for given value of a. f(x) = 5x2- 4x + 7; and a = 4:
SOLUTION
2
F(x) =5x -4x+7
F’(x) =10x-4

F’ (4) =10(4)-4
F’ (a) =36
3- Find f (g(x)) and g (f(x)) for f(x) = 4x + 3; and g(x) = -2x + 1:
SOLUTION
F (G(X)) = f (-2x+1)
= 4(-2x+1) +3
= -8x+7
G (f(x)) = g (4x+3)
= -2(4x+3) +1
=-8x-5
4- Let s be a distance given by
S (t) = 3t3 + 4t2 + 8t:
Find the acceleration.
SOLUTION
3
2
S (t) = 3t +4t +8t
To find the acceleration we derive the function twice
2
S’ (t) = 9t +8t
S’’ (t) = 18t+8
5- Find f0(x) for f(x) = 10/x
Solution
6
F (x) = 10/x =10x
-6
6

F’ (x) = -60x
-7
6- For the function f(x) = 5x2 + 4x. Find f(x + h) - f(x)/h
For:
a. x = 2; h = 0:1.
b. x = 1; h = 0:5.
c. x = 3; h = 1.
Solution
2
a- f (2) =5(2) +4(2) =28
2
F (2+0.1) = 5(2+0.1) +4(2+0.1) =30.09
30.09-28/0.1=20.9
b- F (1) =5(1) +4(1) =9
2
F (1+0.5) =f (1.5) +4(1.5) =17.25
17.25-9/0.5=16.5
2
c- F (3) =5(3) +4(3) =57
2
F (3+1) =5(3+1) +4(3+1) =96
96-57/1=39
7- Find the limit if it exists lim 10x+3
X
3
Lim 10x+3 = lim 10x+3 = 33
+
x 3
x
3
3
8- Differentiate the function y = 5x (2x - 7x)

SOLUTION).
USING CHAIN RULE: F(X).G(X) = F’(X).G(X) + G’(X).F(X)
F(X) =5X
F’(X) =5
3
2
G’(X) = 6X -7
G(X) =2X -7X
3
2
Y’= 5. (2x -7x)+ (6x -7)5x
3
Y’= 4x -7x
2
9- Differentiate the function y= 2x/x +4
F(x) =2x
f’(x) =2
2
G(x) =x +4
g’(x) = 2x
Y’ = g(x).f’(x)-f(x).g’(x)/g(x)
2
2
2
Y’ = (x +4).2 – 2x (2x) / (x +4)
2
2
Y’ = -2(x +8) / (x +4)
2
2
10- Find an expression for dy/dx: y = 3u2 and u = 3x + 1:
Dy/du= 6u,
and
du/dx= 3
Dy/dx= dy/du.U + du/dx.y
2
= 6u. (3x + 1) + 3. (3u )
= 18ux + 6u + 9u
2
2
= 3u + 6ux + 2u
3
2
12- Find the absolute maximum and minimum of the function f(x) = 2x - 3x + 4 on the interval [-1; 2].
SOLUTION

2
F’(x) = 6x -6x
F’(x) =0
2
6x -6x =0
6x(x-1) =0
6x=0
or x-1=0
X=0, X=1
{-1, 0, 1, 2}
F (-1) =-1
F (0) =4
F (1) =3
F (2) =8
The absolute maximum is 8= f (2)
The absolute minimum is -1= f (-1)
3
13- Find dy for given values of x and dx: y = x ;
Dy/dx=3x
2
x = 4 and dx = 0:1
2
dy= 3x .dx
Dy= 3(4). (0.1)
Dy=4.8
3
14- Determine where the function is concave up and down: f(x) = x - 3x + 2:
Concavity can be found using the second derivative.
2
F’ (x) = 3x -3
F’’(x) = 6x
-to find where the function is concave up, we find where the second derivative is positive
6x>0
x>0: the function is concave up for all x- values > 0.
Similarly, the function is concave down for all x-values < 0.

3
2
16- The cost of producing x items is given by: C(x) = x -34x + 5600x: Find the production level that
minimizes the average cost.
3
C(x)/x
2
2
x -34x +5600x/x
= x -34x+5600
3
2
3
17- Find dy=dx by implicit differentiation, given that x + xy + y = 3:
3
2
3
D/dx(x +xy +y ) =d/dx (3)
3
2
3
D/dx(x ) +d/dx (xy ) +d/dx (y ) =0
2
2
2
3x +x.2y.dy/dx+y +3y d/dx=0
2
2
2
2
2
3y d/dx+2yxdy/dx=-3x -y
2
(3y +2yx) dy/dx=-3x -y
2
2
2
Dy/dx= -3x -y /3y +2yx
Find the slope of the tangent line at point (1; 1).
y - y0 = m(x - x0)
18- The cost and the revenue with respect to the number of items is given by C(x) = 200x + 2000 and
2
R(x) = 5x + 100x + 80:
Find the marginal profit
Solution
P(X) = R(x) – C(x)
2
P(x) = (5x +100x+80) – (200x +200)
2
P(x) = 5x – 100x -120
P’(x) = 10x-100