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
9514 1404 393
Answer:
1/2
Step-by-step explanation:
The rules of exponents are helpful.
(a^b)^c = a^(bc)
a^-b = 1/a^b
__

What do you need to do with this?
Answer:
5
Step-by-step explanation:
Step 1:
3x + 4 = 19
Step 2:
3x = 19 - 4
Step 3:
3x = 15
Answer:
x = 5
Hope This Helps :)