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Answers
Hope this helps you !
Using the definition of a derivative:
d
y
d
x
=
lim
h
→
0
f
(
x
+
h
)
−
f
(
x
)
h
, where
h
=
δ
x
We substitute in our function to get:
lim
h
→
0
cos
(
x
+
h
)
−
cos
(
x
)
h
Using the Trig identity:
cos
(
a
+
b
)
=
cos
a
cos
b
−
sin
a
sin
b
,
we get:
lim
h
→
0
(
cos
x
cos
h
−
sin
x
sin
h
)
−
cos
x
h
Factoring out the
cos
x
term, we get:
lim
h
→
0
cos
x
(
cos
h
−
1
)
−
sin
x
sin
h
h
This can be split into 2 fractions:
lim
h
→
0
cos
x
(
cos
h
−
1
)
h
−
sin
x
sin
h
h
Now comes the more difficult part: recognizing known formulas.
The 2 which will be useful here are:
lim
x
→
0
sin
x
x
=
1
, and
lim
x
→
0
cos
x
−
1
x
=
0
Since those identities rely on the variable inside the functions being the same as the one used in the
lim
portion, we can only use these identities on terms using
h
, since that's what our
lim
uses. To work these into our equation, we first need to split our function up a bit more:
lim
h
→
0
cos
x
(
cos
h
−
1
)
h
−
sin
x
sin
h
h
becomes:
lim
h
→
0
cos
x
(
cos
h
−
1
h
)
−
sin
x
(
sin
h
h
)
Using the previously recognized formulas, we now have:
lim
h
→
0
cos
x
(
0
)
−
sin
x
(
1
)
which equals:
lim
h
→
0
(
−
sin
x
)
Since there are no more
h
variables, we can just drop the
lim
h
→
0
, giving us a final answer of:
−
sin
x
.