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DE
Calculus · UNI_FOUNDATION
Differential calculus
Foundation
Core
Stretch
Question 1/8
+12 XP
Differentiate
f
(
x
)
=
2
x
2
.
Secant becomes tangent
f(x) = 2x², x = 1
0
0.5
1
1.5
2
2.5
5
10
x
f(x)
Δx
Δy
(
1
,
2
)
slopes
f
(
x
)
=
2
x
2
Δx
=
1.2
→
m
=
6.4
Δx
=
0.6
→
m
=
5.2
Δx
=
0.2
→
m
=
4.4
Δy
Δx
=
f
′
(
1
)
=
4
curve f
secant Δx = 1.2
secant Δx = 0.6
secant Δx = 0.2
tangent, slope 4
point of tangency
All three secants pass through (1, 2). As Δx shrinks (1.2 → 0.6 → 0.2) the secant slope moves 6.4 → 5.2 → 4.4 — towards the tangent slope f′(1) = 4. That limit is the derivative.
Your answer
Hint 1/3
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