Find derivative of tan square root x by first principle
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Answered by
3
Answer:
f
′
(
x
)
=
1
2
√
x
sec
2
√
x
We note that,
(
1
)
tan
(
A
−
B
)
=
tan
A
−
tan
B
1
+
tan
A
tan
B
(
2
)
lim
θ
→
0
tan
θ
θ
=
1
Explanation:
Here,
f
(
x
)
=
tan
√
x
For first principle method, we take
f
′
(
x
)
=
lim
t
→
x
f
(
t
)
−
f
(
x
)
t
−
x
∴
f
′
(
x
)
=
lim
t
→
x
tan
√
t
−
tan
√
x
t
−
x
=
lim
t
→
x
tan
√
t
−
tan
√
x
1
+
tan
√
t
tan
√
x
t
−
x
×
(
1
+
tan
√
t
tan
√
x
)
=
lim
t
→
x
(
tan
(
√
t
−
√
x
)
t
−
x
)
×
lim
t
→
x
(
1
+
tan
√
t
tan
√
x
)
=
(
1
+
tan
√
x
tan
√
x
)
lim
t
→
x
(
tan
(
√
t
−
√
x
)
√
t
−
√
x
)
1
√
t
+
√
x
=
(
1
+
tan
2
√
x
)
lim
(
√
t
−
√
x
)
→
0
(
tan
(
√
t
−
√
x
)
√
t
−
√
x
)
1
√
t
+
√
x
=
(
sec
2
√
x
)
(
1
)
(
1
√
x
+
√
x
)
f
′
(
x
)
=
1
2
√
x
sec
2
√
x
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