find the derivative of the function sin²x from first principle
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Answered by
0
Step-by-step explanation:
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Answered by
0
Step-by-step explanation:
It is tedious (and space-consuming) to use
lim
h
→
0
on every line, so I hope you'll excuse my approach. We'll simplify the difference quotient first, then find the limit.
(
sin
(
x
+
h
)
)
2
−
(
sin
x
)
2
h
=
(
sin
x
cos
h
+
cos
x
sin
h
)
2
−
(
sin
x
)
2
h
=
sin
2
x
cos
2
h
+
2
sin
x
cos
h
cos
x
sin
h
+
cos
2
x
sin
2
h
−
sin
2
x
h
=
sin
2
x
(
cos
2
h
−
1
)
h
+
2
sin
x
cos
h
cos
x
sin
h
h
+
cos
2
x
sin
2
h
h
=
sin
2
x
cosh
−
1
h
(
cosh
+
1
)
+
2
sin
x
cos
h
cos
x
sin
h
h
+
cos
2
x
sin
h
h
sin
h
Taking limit as
h
→
0
, we get
sin
2
(
0
)
(
0
)
(
cos
0
+
1
)
+
2
sin
x
(
cos
0
)
cos
x
(
1
)
+
cos
0
(
1
)
sin
0
=
0
+
2
sin
x
cos
x
+
0
=
2
sin
x
cos
x
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