Applications of Derivatives

Critical points and extrema

A critical number occurs where $f'(x)=0$ or where the derivative fails to exist while the function is defined. Critical numbers are candidates for local extrema, not automatic maxima or minima.

For $f(x)=x^3-3x^2+2$ , $f'(x)=3x(x-2)$ , so the critical numbers are 0 and 2. Function values are 2 and −2. A sign chart for f′ classifies the first as a local maximum and the second as a local minimum.

First-derivative signs for $f(x)=x^3-3x^2+2$
Interval	Sign of $f'(x)$	Behaviour
$x<0$	Positive	Increasing
$0<x<2$	Negative	Decreasing
$x>2$	Positive	Increasing

The sign changes from positive to negative at 0, giving a local maximum at $(0,2)$ . It changes from negative to positive at 2, giving a local minimum at $(2,-2)$ . These are local conclusions, not global bounds for a cubic on the whole real line.

For absolute extrema on a closed interval, also evaluate the endpoints. The Extreme Value Theorem guarantees attainment of both extrema when the function is continuous on that closed, bounded interval. A zero derivative without a sign change, as for $x^3$ at zero, need not identify an extremum.

Optimization

A rectangle has perimeter 20 metres. Let one side be $x$ metres and the other $y$ metres. Then $2x+2y=20$ , so $y=10-x$ . The area becomes

A(x)=x(10-x),\qquad0<x<10

Differentiate the one-variable objective:

A'(x)=10-2x,\qquad A'(x)=0\Longrightarrow x=5

For $0<x<5$ , the derivative is positive; for $5<x<10$ , it is negative. Area therefore increases to $x=5$ and then decreases. The other side is also 5 metres, and the maximum area is $25\ \mathrm{m}^2$ .

The endpoints 0 and 10 describe collapsed rectangles, not feasible positive-area designs. The area tends to zero at either endpoint. Equivalently, $A(x)=25-(x-5)^2$ makes the upper bound 25 explicit.

This example follows the usual optimization method: establish the objective, apply the constraint, record the domain, find candidates and justify the maximum. The equation $A'(x)=0$ is only one of those steps.

In an idealized expanding circle, radius $r(t)$ and area $A(t)$ both depend on time. Their relationship is $A=\pi r^2$ . Differentiate that relationship before inserting an instant-specific radius:

\frac{dA}{dt}=2\pi r\frac{dr}{dt}

At the instant $r=5$ metres and $dr/dt=1/5$ metre per second,

\left.\frac{dA}{dt}\right|_{r=5,\ dr/dt=1/5}=2\pi(5)\left(\frac15\right)=2\pi\ \mathrm{m^2/s}

Substituting $r=5$ into the original area equation before differentiating would turn the changing radius into a constant and lose the rate information. The positive answer means area is increasing.

The Mean Value Theorem

If f is continuous on [a,b] and differentiable on (a,b), the Mean Value Theorem guarantees at least one point c where the instantaneous rate equals the average rate over the whole interval.

For $f(x)=x^2$ on [1,3], the average slope is 4. Since f′(x)=2x, the theorem point is c=2.

f'(c)=\frac{f(b)-f(a)}{b-a},\qquad a<c<b

The hypotheses require $a<b$ , continuity on $[a,b]$ and differentiability on $(a,b)$ . For $x^2$ on $[1,3]$ , both hold and $2c=4$ gives $c=2$ .

The function $|x|$ on $[-1,1]$ shows why differentiability matters. Its average slope is zero, but its derivatives are −1 on the left and 1 on the right, with no derivative at zero. No interior derivative equals the average slope. The theorem does not apply to that corner.

Concavity and the second derivative

Where $f”(x)>0$ , slopes are increasing and the graph is concave up. Where $f”(x)<0$ , slopes are decreasing and the graph is concave down. A candidate inflection point occurs where concavity can change; checking the sign change matters.

For $f(x)=x^3-3x$ , $f”(x)=6x$ , which changes sign at 0, so x=0 is an inflection point.

For the cubic example, $6x$ is negative on the left of zero and positive on the right, so concavity changes at the point $(0,0)$ . By contrast, $x^4$ has second derivative $12x^2$ , which is zero at zero but never negative. Its concavity does not change there. A zero second derivative is a candidate, not a conclusion.

Motion and interpretation

For a position function s(t), velocity s′(t) gives signed motion and acceleration s″(t) gives the rate of change of velocity. Units and signs matter: negative velocity indicates direction, not “negative speed.”

Where velocity is non-zero, speed increases when velocity and acceleration have the same sign and decreases when they have opposite signs. For example, negative velocity with negative acceleration means the object is moving in the negative direction and speeding up. At a time when velocity is zero, examine the nearby behaviour rather than using that sign shortcut alone.

Practice questions

A rectangle has perimeter 20 metres. Which sides maximize its area?
For $f(x)=x^2$ on $[1,3]$ , find a point where derivative equals average slope.
If $A=\pi r^2$ , $r=5$ metres and $dr/dt=1/5$ metre per second, find $dA/dt$ .

Check your answers

Write the sides as $x$ and $10-x$ . Then $A=x(10-x)$ and $A’=10-2x$ . The derivative changes from positive to negative at $x=5$ , giving sides 5 m and 5 m and area $25\ \mathrm{m^2}$ .
The average slope is $(9-1)/(3-1)=4$ . Since $f'(c)=2c$ , choose $c=2\in(1,3)$ .
Differentiate before substituting: $dA/dt=2\pi r(dr/dt)$ . At the given instant this is $2\pi(5)(1/5)=2\pi$ square metres per second.

Sources and further reading

These references support the definitions, theorem assumptions and standard conventions used in this article.

OpenStax Calculus Volume 1: Linear Approximations and Differentials. Local derivative applications.
OpenStax Calculus Volume 1: Related Rates. Rates connected by equations.
OpenStax Calculus Volume 1: Applied Optimization Problems. Optimization with derivatives.
OpenStax Calculus Volume 1: The Mean Value Theorem. Average and instantaneous rates.