01 Key Concepts
Rates of Change
The derivative of a quantity with respect to time gives its rate of change, such as velocity (rate of change of position).
Velocity and Acceleration
If s(t) is position, then velocity v(t) = s'(t), and acceleration a(t) = v'(t) = s''(t).
Increasing/Decreasing Functions
If f'(x) > 0 on an interval, f is increasing there. If f'(x) < 0, f is decreasing.
Concavity
If f''(x) > 0, the graph is concave up (like a cup). If f''(x) < 0, it is concave down (like a frown).
Critical Points
Points where f'(x) = 0 or f'(x) is undefined — candidates for local maxima or minima.
02 Key Formulas
- v(t) = s'(t)
- a(t) = v'(t) = s''(t)
03 Solved Examples
Example 1 A particle's position is s(t) = t^2 - 4t. Find its velocity at t = 3.
- Velocity is the derivative of position: v(t) = s'(t) = 2t - 4.
- Substitute t = 3: 2(3) - 4.
Answer: v(3) = 2
Example 2 For f(x) = x^2 - 6x, find the interval where f is increasing.
- Find f'(x) = 2x - 6.
- Set f'(x) > 0: 2x - 6 > 0, so x > 3.
Answer: f is increasing for x > 3
Example 3 Find the critical points of f(x) = x^3 - 3x.
- Find f'(x) = 3x^2 - 3.
- Set f'(x) = 0: 3x^2 - 3 = 0, so x^2 = 1, giving x = 1 or x = -1.
Answer: Critical points at x = 1 and x = -1
04 Practice Questions
1If s(t) = 3t^2, find velocity at t=2.
12
2If v(t) = 4t - 1, find acceleration.
4
3For f(x) = x^2 - 4x, find where f is decreasing.
x < 2
4Find critical points of f(x) = x^2 - 8x.
x = 4
5If f''(x) > 0 on an interval, describe the concavity.
Concave up
📄 Applications of Derivatives — Downloadable Worksheet
10 questions with a full answer key. Grab the PDF to print, or try the interactive version in your browser.