differentiate from 1st priciple f(x)=√3x+4
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Answered by
1
step by step explanation
Answered by
0
Answer:
f
'
(
x
)
=
1
2
√
x
+
3
Explanation:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
f
(
x
)
=
√
x
+
3
,
f
(
x
+
h
)
=
√
x
+
h
+
3
, then
f
'
(
x
)
=
lim
h
→
0
√
x
+
h
+
3
−
√
x
+
3
h
If we evaluate this right away, we get
lim
h
→
0
√
x
+
h
+
3
−
√
x
+
3
h
=
√
x
+
3
−
√
x
+
3
0
=
0
0
,
so we need to simplify as this is an indeterminate form.
Multiply the entire limit by the numerator's conjugate, which is
√
x
+
h
+
3
+
√
x
+
3
√
x
+
h
+
3
+
√
x
+
3
. This is the same as multiplying by
1
.
f
'
(
x
)
=
lim
h
→
0
√
x
+
h
+
3
−
√
x
+
3
h
⋅
√
x
+
h
+
3
+
√
x
+
3
√
x
+
h
+
3
+
√
x
+
3
The numerator becomes
√
x
+
h
+
3
−
√
x
+
3
⋅
[
√
x
+
h
+
3
+
√
x
+
3
]
=
x
+
h
+
3
−
(
x
+
3
)
=
x
+
h
+
3
−
x
−
3
=
h
f
'
(
x
)
=
lim
h
→
0
x
+
h
+
3
−
x
−
3
h
(
√
x
+
h
+
3
+
√
x
+
3
)
f
'
(
x
)
=
lim
h
→
0
h
h
(
√
x
+
h
+
3
+
√
x
+
3
)
f
'
(
x
)
=
lim
h
→
0
1
√
x
+
h
+
3
+
√
x
+
3
f
'
(
x
)
=
1
√
x
+
3
+
√
x
+
3
f
'
(
x
)
=
1
2
√
x
+
3
shifarmaan:
can't understand a single thing:(
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