Derivative of √ sinx
Answers
Answer:
Answer:
d
d
x
√
sin
x
=
cos
x
2
√
sin
x
Explanation:
Using the limit definition of the derivative we have:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
So for the given function, where
f
(
x
)
=
√
sin
x
, we have:
f
'
(
x
)
=
lim
h
→
0
√
sin
(
x
+
h
)
−
√
sin
x
h
=
lim
h
→
0
√
sin
(
x
+
h
)
−
√
sin
x
h
⋅
√
sin
(
x
+
h
)
+
√
sin
x
√
sin
(
x
+
h
)
+
√
sin
x
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
(
√
sin
(
x
+
h
)
+
√
sin
x
)
Then we can use the trigonometric identity:
sin
(
A
+
B
)
≡
sin
A
cos
B
+
cos
A
sin
B
Giving us:
f
'
(
x
)
=
lim
h
→
0
sin
x
cos
h
+
cos
x
sin
h
−
sin
x
h
(
√
sin
(
x
+
h
)
+
√
sin
x
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
+
cos
x
sin
h
h
(
√
sin
(
x
+
h
)
+
√
sin
x
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
h
(
√
sin
(
x
+
h
)
+
√
sin
x
)
+
cos
x
sin
h
h
(
√
sin
(
x
+
h
)
+
√
sin
x
)
=
lim
h
→
0
cos
h
−
1
h
sin
x
√
sin
(
x
+
h
)
+
√
sin
x
+
sin
h
h
cos
x
√
sin
(
x
+
h
)
+
√
sin
x
Then we use two very standard calculus limits:
lim
θ
→
0
sin
θ
θ
=
1
, and
lim
θ
→
0
cos
θ
−
1
θ
=
0
, and #
And we can now evaluate the limits:
f
'
(
x
)
=
0
×
sin
x
√
sin
(
x
)
+
√
sin
x
+
1
×
cos
x
√
sin
(
x
)
+
√
sin
x
=
cos
x
2
√
sin
(
x
)
Answer:
1/2√sinx x cosx is the answer