Applications of Derivatives — Maxima, Minima & Rate of Change

Applications of derivatives come down to two core ideas: using $f'(x)$ to measure an instantaneous rate of change, and using $f'(x)$ to find where a quantity reaches a maximum or minimum. If a problem asks how fast something is changing right now, or when a value is highest or lowest, the derivative is usually the tool you need.

When to use this approach

Reach for derivatives when the question is about a rate or an extreme value. The derivative $f'(x)$ tells you how the output changes as the input changes at a specific point, and that single idea covers several common tasks:

• In graph terms, it gives the slope of the tangent line.
• In motion, it can give velocity as the rate of change of position.
• In optimization, it locates where a quantity stops increasing and starts decreasing, or the reverse.

The context shifts across geometry, physics, economics, and engineering, but the core question stays the same: how does one quantity respond when another changes? One condition matters from the start: if the question asks for an absolute maximum or minimum on a closed interval, you must compare critical points and endpoints. Checking only where $f'(x)=0$ is not enough.

The procedure, step by step

1. Differentiate the function. Find $f'(x)$ and connect it to the meaning of the problem, such as slope, speed, or rate of profit change.
2. Find critical points. Solve $f'(x)=0$, and also check where $f'(x)$ does not exist while $f(x)$ is still defined.
3. Classify the result. Use the sign of $f'(x)$ around the point, or when it applies the sign of $f''(x)$, to decide maximum, minimum, or neither.
4. Check endpoints when needed. For an absolute maximum or minimum on a closed interval, compare the endpoints as well.

For Step 3, the sign tests are: $f'(x)$ changing from positive to negative gives a local maximum; from negative to positive gives a local minimum. If the function is twice differentiable near the point, $f''(x) < 0$ suggests concave down and a local maximum, while $f''(x) > 0$ suggests concave up and a local minimum. The second-derivative shortcut only helps when it exists and is not $0$ at the critical point; if $f''(x)=0$, use another test. Remember that not every critical point is an extremum, because a function can flatten out and keep moving in the same overall direction.

A full example through every step

Suppose the height of a ball is modeled by

where $h$ is in meters and $t$ is in seconds. This model is useful only over the time interval where the ball is actually in flight.

Step 1, differentiate:

This is the instantaneous rate of change of height with time, the vertical velocity in meters per second.

Step 2, find the critical point:

At $t=2$, the vertical velocity is $0$, making it a candidate.

Step 3, classify: before $t=2$ the derivative is positive, so height is increasing; after $t=2$ it is negative, so height is decreasing. That sign change means a maximum at $t=2$.

Evaluate the original function to get the actual value:

So the maximum height is $22$ meters. The example shows both main applications at once: $h'(t)$ gives a rate of change, and setting $h'(t)=0$ helps find when the height is greatest.

Where each step traps students

• At Step 2 to 3: treating every solution of $f'(x)=0$ as a maximum or minimum without checking what happens around it.
• At Step 4: forgetting endpoints when the question asks for an absolute extremum on a closed interval.
• Across steps: mixing up the original function and its derivative. The derivative tells you where the extremum happens, but the actual value comes from the original function.
• Units: if position is in meters and time in seconds, the derivative is in meters per second.
• Domain: using a model outside the interval where it makes physical sense.

In early calculus, derivatives are used for curve sketching, tangent lines, motion, and optimization. In science and engineering they describe changing systems such as velocity, acceleration, heat flow, current, or growth. In economics they describe marginal cost or marginal revenue, which are rates of change as well.

Work one through yourself

Take

Run the steps: find $f'(x)$, locate the critical point, decide maximum or minimum from the sign change, then compute the function value there. For a case where two changing quantities are linked, the page on related rates shows how the same derivative idea extends.