Applications of Derivatives

The One Thing

Applications of derivatives use the value and sign of derivatives to analyse the behaviour of functions. Derivatives determine increasing and decreasing intervals, construct tangents and normals, and identify and classify local and absolute extrema, including in optimization problems.

Definitions and Results

• Derivative: For a function $y=f(x)$, the derivative $f^{\prime }(x)$ represents the instantaneous rate of change of $y$ with respect to $x$ and the slope of the tangent to the curve. At $x=a$,

$$f^{\prime }(a)=\underset{h\to 0}{\lim }\frac{f(a+h)-f(a)}{h},$$ when the limit exists.

• Increasing Function: A function is increasing on an interval if its values increase as $x$ increases. If $f^{\prime }(x)>0$ throughout the interval, then the function is strictly increasing there.

• Decreasing Function: A function is decreasing on an interval if its values decrease as $x$ increases. If $f^{\prime }(x)<0$ throughout the interval, then the function is strictly decreasing there.

• Critical Point: A point in the domain where $f^{\prime }(x)=0$ or where $f^{\prime }(x)$ does not exist, provided that the function itself is defined there.

• Stationary Point: A point where $f^{\prime }(x)=0$. A stationary point may be a local maximum, a local minimum, or neither.

• Tangent: A line that touches a curve at a point and has the same slope as the curve at that point. For $y=f(x)$, the tangent slope at $x=a$ is

$$m_t=f^{\prime }(a).$$

• Normal: A line perpendicular to the tangent at the point of contact. When $f^{\prime }(a)\ne 0$, its slope is

$$m_n=-\frac{1}{f^{\prime }(a)}.$$

• Local Maximum: The function has a local maximum at $x=a$ if $f(a)$ is greater than or equal to nearby function values. Usually, $f^{\prime }(x)$ changes from positive to negative at $a$.

• Local Minimum: The function has a local minimum at $x=a$ if $f(a)$ is less than or equal to nearby function values. Usually, $f^{\prime }(x)$ changes from negative to positive at $a$.

• Absolute Maximum: The greatest value of a function over its entire specified domain or interval.

• Absolute Minimum: The least value of a function over its entire specified domain or interval.

• First Derivative Test: A critical point is classified by the sign change of $f^{\prime }(x)$:

- positive to negative: local maximum; - negative to positive: local minimum; - no sign change: generally neither.

• Second Derivative Test: If $f^{\prime }(a)=0$, then:

- $f^{\prime \prime }(a)<0$ implies a local maximum at $x=a$; - $f^{\prime \prime }(a)>0$ implies a local minimum at $x=a$; - $f^{\prime \prime }(a)=0$ gives no definite conclusion.

• Optimization: The process of finding the greatest or least possible value of a quantity using mathematical conditions and derivatives.

• Interior extrema condition: At a local maximum or minimum occurring at an interior point, $f^{\prime }(a)=0$ or $f^{\prime }(a)$ does not exist. This condition alone does not prove that a maximum or minimum occurs.

• Absolute extrema on a closed interval: For a function on $[a,b]$, evaluate the function at every critical point inside the interval and at both endpoints $a$ and $b$, then compare the values.

Worked Methods

1. Determining increasing and decreasing intervals

• Find $f^{\prime }(x)$.
• Find the critical points by solving $f^{\prime }(x)=0$ and identifying points where $f^{\prime }(x)$ does not exist, provided $f(x)$ is defined there.
• Use the critical points to divide the domain into intervals.
• Test the sign of $f^{\prime }(x)$ in each interval.
• State the intervals:

- $f^{\prime }(x)>0$: the function is increasing; - $f^{\prime }(x)<0$: the function is decreasing.

The sign of the first derivative gives the direction of change of the function.

2. Finding the equation of a tangent

For the curve $y=f(x)$ at $(a,f(a))$:

• Calculate $f^{\prime }(x)$.
• Evaluate the derivative at $x=a$ to obtain the tangent slope:

$$m_t=f^{\prime }(a).$$

• Substitute the point and slope into the point-slope equation:

$$y-f(a)=f^{\prime }(a)(x-a).$$

Thus, the tangent at $(a,f(a))$ is $$y-f(a)=f^{\prime }(a)(x-a).$$

If $f^{\prime }(a)=0$, the tangent is horizontal and its equation is $$y=f(a).$$

3. Finding the equation of a normal

For the curve $y=f(x)$ at $(a,f(a))$:

• Calculate the tangent slope $f^{\prime }(a)$.
• If $f^{\prime }(a)\ne 0$, calculate the normal slope using the negative reciprocal:

$$m_n=-\frac{1}{f^{\prime }(a)}.$$

• Use the point-slope equation:

$$y-f(a)=-\frac{1}{f^{\prime }(a)}(x-a).$$

The negative reciprocal is required because perpendicular non-vertical lines have slopes whose product is $-1$.

If the tangent is horizontal, then $f^{\prime }(a)=0$. The tangent is $$y=f(a),$$ and the normal is the vertical line $$x=a.$$

4. Classifying stationary points using the first derivative

• Find $f^{\prime }(x)$.
• Solve $f^{\prime }(x)=0$ to locate stationary points.
• Examine the sign of $f^{\prime }(x)$ immediately to the left and right of each point.
• Apply the sign-change criteria:

- $+$ to $-$: local maximum; - $-$ to $+$: local minimum; - no sign change: generally neither.

A stationary point may therefore be neither a maximum nor a minimum, such as a point where the graph continues increasing on both sides.

5. Classifying stationary points using the second derivative

• Find $f^{\prime }(x)$ and solve $f^{\prime }(a)=0$.
• Find $f^{\prime \prime }(x)$.
• Evaluate $f^{\prime \prime }(a)$.
• Apply the test:

- $f^{\prime \prime }(a)<0$: local maximum; - $f^{\prime \prime }(a)>0$: local minimum; - $f^{\prime \prime }(a)=0$: no definite conclusion.

The second derivative test requires $f^{\prime }(a)=0$.

6. Finding absolute maxima and minima on a closed interval

For a function on $[a,b]$:

• Find all critical points in the interval.
• Evaluate the function at every critical point inside the interval.
• Evaluate the function at the endpoints $a$ and $b$.
• Compare all resulting values.
• The greatest value is the absolute maximum, and the least value is the absolute minimum.

Critical points alone are insufficient because an absolute extremum may occur at a boundary point.

7. Solving an optimization problem

• Express the required quantity as a function of one variable.
• Determine the feasible domain of that variable.
• Differentiate the function.
• Find critical points by solving $f^{\prime }(x)=0$ and considering where $f^{\prime }(x)$ does not exist.
• Compare the relevant function values, including boundary values where required.
• State the result with appropriate units.

This method converts practical questions about greatest profit, least material, shortest distance, or largest area into mathematical optimization problems.

Where It Goes Wrong

• Treating $f^{\prime }(a)=0$ as proof of a maximum or minimum; it is only a necessary condition for an interior differentiable extremum.
• Forgetting to test intervals on both sides of critical points when applying the first derivative test.
• Assuming every stationary point is an extremum; a stationary point may have no sign change and be neither a maximum nor a minimum.
• Applying the second derivative test when $f^{\prime \prime }(a)=0$, even though this gives no definite conclusion.
• Finding absolute extrema using critical points alone and omitting the endpoints of a closed interval $[a,b]$.
• Using the negative reciprocal for a normal without checking the tangent condition: when $f^{\prime }(a)=0$, the normal is $x=a$, not a line with slope $-1/f^{\prime }(a)$.

What Gets Asked

• Define the derivative and interpret it as an instantaneous rate of change and the slope of a tangent.
• Use the sign of $f^{\prime }(x)$ to determine intervals on which a function is increasing or decreasing.
• Find critical and stationary points.
• Determine the equation of a tangent at $(a,f(a))$.
• Determine the equation of a normal, including the special case of a horizontal tangent.
• Classify stationary points using the first derivative test.
• Classify stationary points using the second derivative test.
• Determine absolute maxima and minima on a closed interval by comparing critical-point and endpoint values.
• Explain why a stationary point need not be a maximum or minimum.
• Formulate and solve optimization problems involving greatest profit, least material, shortest distance, or largest area, and state answers with appropriate units.