From a Function to Its Second Derivative
A function describes how one quantity depends on another. Its first derivative describes the function’s instantaneous rate of change. The second derivative asks a further question: how quickly is that rate changing?
This idea appears in many settings. If a function gives position as time passes, its first derivative describes the rate at which position changes. The second derivative describes how that rate changes over time. In any setting, the second derivative is a new function, with its own values, meaning, and units. It is not the original function, and it is not the square of the first derivative.
The notation \(f''\) is read “\(f\) double prime.” If \(y=f(x)\), the second derivative can also be written \(\frac{d^2y}{dx^2}\). The exponent \(2\) indicates that differentiation has been performed twice; it does not mean that the first derivative has been squared.
Why It Measures a Rate of Change
The definition of a derivative makes the interpretation precise. The derivative of a function \(g\) at \(x\), when it exists, is the limit of the average rates of change of \(g\) over intervals starting at \(x\). Apply that definition to the function \(g=f'\). Its average rate of change compares two values of the original function’s first derivative.
This limit says that \(f''(x)\) is the instantaneous rate at which \(f'\) changes as the input moves through \(x\). For a small change \(h\), the quotient measures the average change in \(f'\) per unit change in the input. As \(h\) approaches zero, that average rate approaches the instantaneous rate, if the limit exists.
That last condition matters. A function may have a first derivative at a point even though its first derivative is not differentiable there. In that case, the second derivative at that point does not exist.
Worked Example: Three Values for One Polynomial
Worked Example: Compare the Function and Its Derivatives
Let \(f(x)=2x^3-3x^2+4x-1\). Find \(f(1)\), \(f'(1)\), and \(f''(1)\).
Step 1: Find the function value.
Step 2: Differentiate once. Using the basic power rule and differentiating each term,
Therefore,
Step 3: Differentiate the first derivative.
Check: Use the derivative definition on \(f'\) at \(x=1\):
As \(h\to 0\), this quotient approaches \(6\), confirming \(f''(1)=6\). At the same input, the three values are \(2\), \(4\), and \(6\); each belongs to a different function in the sequence \(f\), \(f'\), and \(f''\).
Units of the Second Derivative
Units provide a useful way to distinguish the function and its derivatives. If \(f\) measures a quantity in units \(U\) and \(x\) is measured in units \(X\), then \(f'(x)\) has units \(U/X\): it measures change in \(f\) per unit change in \(x\). Differentiating once more divides by another unit of \(X\), so \(f''(x)\) has units \(U/X^2\).
This follows directly from the difference quotient for \(f''\): the numerator is a change in \(f'\), measured in \(U/X\), and the denominator is a change in \(x\), measured in \(X\). Dividing gives \(U/X^2\). Units can therefore help catch an interpretation or calculation that confuses the first and second derivatives.
Worked Example: Position and Change in Its Rate
Worked Example: Read the Derivatives of a Position Function
A particle’s position, in meters, is given as a function of time \(t\), in seconds:
Find its position, its rate of change of position, and the rate of change of that rate at \(t=1\text{ s}\).
Step 1: Evaluate the position. Substitute \(t=1\text{ s}\), keeping the units in each term:
Step 2: Differentiate to find the rate of change of position.
Each term in \(s'(t)\) has units of meters per second. At \(t=1\text{ s}\),
Step 3: Differentiate again. The second derivative gives the rate of change of \(s'(t)\):
At \(t=1\text{ s}\),
Check: Differentiate the displayed expression for \(s'(t)\): the derivative of its first term is \(\left(6\frac{\text{m}}{\text{s}^3}\right)t\), the derivative of its second term is \(-12\frac{\text{m}}{\text{s}^2}\), and the constant term has derivative \(0\). This agrees with \(s''(t)\). At one second, the position is \(6\) meters, its rate of change is \(0\) meters per second, and the rate of change of that rate is \(-6\) meters per second squared. These are different quantities with different units.
A Second Derivative Can Be Zero While the Other Values Are Not
A second derivative equal to zero means that the instantaneous rate of change of the first derivative is zero at that input. It does not mean that the function itself or its first derivative must be zero there.
Worked Example: A Zero Second Derivative
Let \(g(x)=x^3-3x^2+2x\). Find the function and derivative values at \(x=1\), and interpret the second derivative value.
Step 1: Evaluate the function.
Step 2: Find and evaluate the first derivative.
Step 3: Find and evaluate the second derivative.
Check: The derivative quotient for \(g'\) at \(1\) is
Its limit as \(h\to 0\) is \(0\), confirming the second derivative. Here \(g(1)=0\), \(g'(1)=-1\), and \(g''(1)=0\). In particular, the second derivative’s value is not the value of the function or its first derivative.
When the Second Derivative Does Not Exist
To differentiate a function twice at a point, it is not enough that the original function has a first derivative at that point. The first derivative itself must have a derivative there.
Worked Example: A First Derivative Without a Second Derivative at Zero
Consider \(q(x)=x|x|\). For \(x\ge 0\), \(q(x)=x^2\); for \(x<0\), \(q(x)=-x^2\). At \(x=0\), the derivative of \(q\) is
For \(x>0\), \(q'(x)=2x\), and for \(x<0\), \(q'(x)=-2x\). Together with \(q'(0)=0\), this says \(q'(x)=2|x|\) for all \(x\). Now test whether \(q'\) has a derivative at zero:
For \(h>0\), this quotient is \(2\); for \(h<0\), it is \(-2\). The two one-sided limits differ, so the limit does not exist. Thus \(q'(0)\) exists, but \(q''(0)\) does not.
A Common Notation Mistake
The second derivative is found by differentiating the first derivative. It is not found by squaring the first derivative.
The correction is to differentiate \(f'(x)\):
For a numerical check at \(x=2\), the incorrect expression gives \(9(2)^4=144\), while the correct derivative gives \(6(2)=12\). The derivative definition confirms the correct value:
Another useful check is to compare units: if \(x\) is measured in seconds and \(f\) in meters, the second derivative must have units of meters per second squared, not the squared units of \(f'\).
Why the Second Derivative Matters
The second derivative adds information that the function value and first derivative do not provide by themselves. In a position model, the function gives location, the first derivative gives the rate at which location changes, and the second derivative gives the rate at which that first rate changes. In other applications, the same structure describes how a changing rate itself changes.
For interpretation, keep three questions separate: What quantity does the original function measure? What rate does its first derivative measure? What rate of change does the second derivative measure? State units at each stage. This prevents a numerical value from being assigned the wrong meaning.
Check Your Understanding
Use the definitions and interpretations in this tutorial to answer each question.
- In words, what does \(f''(x)\) measure, and which function is being differentiated to obtain it?
- If \(f\) is measured in liters and \(x\) in minutes, what units does \(f''(x)\) have?
- For \(p(x)=x^3-2x^2\), find \(p'(x)\), \(p''(x)\), and \(p''(1)\).
- A position function is measured in meters and its input is time in seconds. What are the units of its first and second derivatives?
- Can a first derivative exist at a point while the second derivative fails to exist there? Explain how the example \(q(x)=x|x|\) demonstrates this.
