differentiate sinx^0
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Answered by
1
Answer:
Answer:
d
d
x
sin
x
=
cos
x
Explanation:
By definition of the derivative:
f
'
(
x
)
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
So with
f
(
x
)
=
sin
x
we have;
f
'
(
x
)
=
lim
h
→
0
sin
(
x
+
h
)
−
sin
x
h
Using
sin
(
A
+
B
)
=
sin
A
cos
B
+
sin
B
cos
A
we get
f
'
(
x
)
=
lim
h
→
0
sin
x
cos
h
+
sin
h
cos
x
−
sin
x
h
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
+
sin
h
cos
x
h
=
lim
h
→
0
(
sin
x
(
cos
h
−
1
)
h
+
sin
h
cos
x
h
)
=
lim
h
→
0
sin
x
(
cos
h
−
1
)
h
+
lim
h
→
0
sin
h
cos
x
h
=
(
sin
x
)
lim
h
→
0
cos
h
−
1
h
+
(
cos
x
)
lim
h
→
0
sin
h
h
We know have to rely on some standard limits:
lim
h
→
0
sin
h
h
=
1
, and
lim
h
→
0
cos
h
−
1
h
=
0
And so using these we have:
f
'
(
x
)
=
0
+
(
cos
x
)
(
1
)
=
cos
x
Hence,
d
d
x
sin
x
=
cos
xtep-by-step explanation:
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Answered by
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