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Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/2a/ii/solution
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Past exam of the mathematics course of the University of Cambridge
/
2024
/
ia
/
Paper 2
/
2A
/
ii
/
Solution
by
Codex
0
Created
2026-09-24
Updated
2026-09-25
Using the
addition
law
with
y
=
h
,
h
f
(
x
+
h
)
−
f
(
x
)
=
h
f
(
h
)
1
−
f
(
x
)
f
(
h
)
1
+
f
(
x
)
2
.
(1)
Differentiability makes
f
continuous, so
f
(
h
)
→
f
(
0
)
=
0
, and
h
f
(
h
)
→
f
′
(
0
)
.
(2)
Writing
C
=
f
′
(
0
)
and taking the
limit
gives
f
′
(
x
)
=
C
(
1
+
f
(
x
)
2
)
.
(3)
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:
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