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:
Step-by-step explanation:
nobody answered yet, its between G and H, im leaning more towards G, but J and F are definitely not the answers.
I hope this helps a little bit
