Econ 100
Exam 4 Maths practice worksheet solutions
Solve for y
1. y =
2−5 3−4
2−6 3−3
2. y =
10−3 3−10
10−6 3−6
=
2
3
=
103
34
or =
1000
81
3. y =
5−3 3−5
5−6 36
=
4. y =
2−2 5−3
5−6 2−2
= 53 or 125
53
311
or
dC
Derive MPC ( dY
) and APC ( YC ) for the following:
Solve for loge (y):
1. y 6 = x3 ⇒ loge (y) =
1
2
loge (x)
2. y −3 = x−4 ⇒ loge (y) =
3. y 7 = x2 ⇒ loge (y) =
2
7
4
3
loge (x)
loge (x)
1. C = 10 + 0.6Y ⇒
dC
dY
= 0.6, C/Y =
2. C = 70+0.75Y ⇒
dC
dY
= 0.75, C/Y =
3. C = 10 + Y ⇒
Derive
dy
dx
= −0.5e−0.5x + 2x
2. y = 15 + loge x + 3x ⇒
dy
dx
=
∂y
∂x
1
x
3. y = 32 + loge x + x4 ⇒
=
1
x
10
Y
70
Y +0.75
+1
dy
dx
+ 4x3
= x1 +5x4 +12x3
∂y
∂x
5. y = 12ez loge x + x5 loge z ⇒
5x4 loge z
= 27z + 5x4
3. y = 32z 2 + 18zx + 12x2 z 2 + 10x ⇒
18z + 24xz 2 + 10
∂y
∂x
4. y = 10xz + ex loge z + 13 loge z ⇒
10z + ex loge z
∂y
∂x
∂y
∂L
=
dy
dx
4. y = 72+loge x+x5 +3x4 ⇒
+3
= 15z + 5x4 z 2
2. y = 25z + 27xz + x5 + z 2 ⇒
1−b
1−b(1−t)
= 1, , C/Y =
+ 0.6
∂y
∂x :
1. y = 15xz + x5 z 2 ⇒
Find
dC
dY
10
Y
dy
dx :
1. y = e−0.5x + x2 ⇒
Derive
125
177147
∂y
∂G :
1−b
1−b(1−t)
+
y =
a
1−b(1−t)
−
bL
1−b(1−t)
=
12ez
x
+
∂y
∂x
=
6. y = 15ez + 12ex + x loge z + z loge x ⇒
12ex + loge z + xz
=
7. y = e0.8x + e0.2z ⇒
+
∂y
∂x
∂y
∂x
= 0.8e0.8x
8. y = 5ez + 3e0.9x + x2 loge z + 2z 3 loge (x2 ) +
3
∂y
= 2.7e0.9x +2x loge z+ 4zx +5x4 z −2
x5 z −2 ⇒ ∂x
=
e−f r
1−b(1−t)
1
+
G
1−b(1−t)
⇒
∂y
∂L
+
∂y
∂G
b
= − 1−b(1−t)
+
1
1−b(1−t)
=