∑
Math Atelier
Home
Learn
Practice
Assessments
Worksheets
Videos
DE
Start practice
Home
→
Learn
→
Practice
→
Assessments
→
Worksheets
→
Videos
→
Start practice
DE
Calculus · SEC2_GYM
Differentiation rules
Foundation
Core
Stretch
Question 1/8
+12 XP
Differentiate
f
(
x
)
=
5
x
3
.
Secant becomes tangent
f(x) = 5x³, x = 1
0
0.5
1
1.5
10
20
30
x
f(x)
Δx
Δy
(
1
,
5
)
slopes
f
(
x
)
=
5
x
3
Δx
=
0.6
→
m
=
25.8
Δx
=
0.3
→
m
=
19.95
Δx
=
0.1
→
m
=
16.55
Δy
Δx
=
f
′
(
1
)
=
15
curve f
secant Δx = 0.6
secant Δx = 0.3
secant Δx = 0.1
tangent, slope 15
point of tangency
All three secants pass through (1, 5). As Δx shrinks (0.6 → 0.3 → 0.1) the secant slope moves 25.8 → 19.95 → 16.55 — towards the tangent slope f′(1) = 15. That limit is the derivative.
Your answer
Hint 1/3
Check