f
'
(
x
)
=
1
(
x
+
1
)
2
Explanation:
differentiating from first principles
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
f
'
(
x
)
=
lim
h
→
0
x
+
h
x
+
h
+
1
−
x
x
+
1
h
the aim now is to eliminate h from the denominator
f
'
(
x
)
=
lim
h
=0
(
x
+
h
)
(
x
+
1
)−
x
(
x
+
h
+
1)
h
(
x
+
1
)
(
x
+
h
+
1
)
f
'
(
x
)
=
lim
h
→
0
x
2
+
h
x
+
x
+
h
−
x
2
−
h
x
−
x
h
(
x
+
1
)
(
x+h
+
1
)
f
'
(
x
)
=
lim
h
→
0
h
1
h
1
(
x
+
1
)
(
x
+
h
+1
)
f
'
(
x
)
=
1
(
x
+
1
)
2
Answer:
C: B+C
Step-by-step explanation:
A-B is even so A and B are either both even or odd.
A-C is odd so one is even and the other is odd.
Because A is common to both cases B and C have to be of different parities.
In answer A, B, and D we are not sure of the parity of the numbers.
However, in answer C we know that B and C are of different parities, and and even plus and odd is odd.
When you rotate 90 degrees clockwise, (x,y) becomes (y,-x).
Therefore:
G = (3,1)
H = (0, 4)
I = (-2, -3)
Answer:
I think it would be B
Step-by-step explanation:
