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Derivative Functions and Interpretation · Tutorial 376 of 1495

Derivative Functions As Input-Dependent Rates

Understand the derivative as a function of the input, rather than as one fixed number.

Intermediate 9 min read

What You'll Learn

  • Define the derivative function on the set of inputs where the derivative exists
  • Distinguish the function \(f'\) from a particular value \(f'(a)\)
  • Derive derivative functions from difference quotients while keeping the input variable
  • Recognize when a derivative function is constant or depends on its input
  • Identify why a derivative function may not be defined everywhere on the original domain

From One Rate to a Function of Rates

A derivative is often introduced as a number: the instantaneous rate of change of a function at a particular input. But the input itself can vary. At one input the rate may have one value; at another input it may have a different value. Recording the rate at every input where it exists creates a new function.

This distinction is useful because a derivative is not generally one fixed number. The notation \(f'(x)\) describes a rule whose input is \(x\). By contrast, \(f'(a)\) is the particular output of that rule when its input is the specified number \(a\).

Central idea: The derivative function \(f'\) assigns to each input \(x\) where \(f\) is differentiable the instantaneous rate of change of \(f\) at that input.

Definition and Domain

Let \(f:I\to\mathbb{R}\), where \(I\) is an open interval. At an input \(x\in I\), the derivative is defined by the limit of the difference quotient, if that limit exists as a finite real number:

$$ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. $$

Here \(h\) is the change in input. In taking the limit, \(x\) is held fixed and \(h\) approaches zero. The value of the limit may depend on which \(x\) was held fixed.

Definition: The derivative function of \(f\) is the function \(f':D\to\mathbb{R}\) given by the derivative limit, where $$ D=\left\{x\in I:\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\text{ exists as a finite real number}\right\}. $$ Thus the domain of \(f'\) consists precisely of the inputs in \(I\) where \(f\) is differentiable.

The domain of the derivative function need not be all of \(I\). If the derivative limit fails at some input, \(f'\) has no value there. This does not prevent \(f'\) from being a function: a function can have a domain that is smaller than the domain of the original function.

It helps to keep the roles of the two functions separate. The original function \(f\) assigns an output \(f(x)\) to an input \(x\). The derivative function \(f'\) assigns a rate \(f'(x)\) to that same input, where the derivative exists.

NotationWhat it denotes
\(f\)The original function
\(f(x)\)The output of the original function at input \(x\)
\(f'\)The derivative function
\(f'(x)\)The output of the derivative function at input \(x\)

Worked Example: Deriving the Function for a Square

Worked Example: Deriving the Function for a Square

Let \(f:\mathbb{R}\to\mathbb{R}\) be \(f(x)=x^2\). Find the derivative function from its definition.

Begin with an arbitrary fixed input \(x\), rather than choosing one numerical input. For \(h\ne 0\), the difference quotient is

$$ \begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{(x+h)^2-x^2}{h}\\ &=\frac{x^2+2xh+h^2-x^2}{h}\\ &=\frac{2xh+h^2}{h}\\ &=2x+h. \end{aligned} $$

Now take the limit as \(h\to 0\), with \(x\) fixed:

$$ f'(x)=\lim_{h\to 0}(2x+h)=2x. $$

The calculation works for every real \(x\), so the derivative function is \(f':\mathbb{R}\to\mathbb{R}\), \(f'(x)=2x\). As a check, substituting any fixed input \(a\) into the original difference quotient gives \(2a+h\), whose limit is \(2a\), agreeing with the formula \(f'(a)=2a\).

The expression \(2x\) is not one derivative value. It is the rule for producing a derivative value from an input. For example, substituting \(x=3\) into this rule gives \(f'(3)=6\). The rule \(f'(x)=2x\) describes all such outputs, not just the one obtained at \(3\).

Worked Example: A Linear Function Has a Constant Derivative Function

Worked Example: A Linear Function Has a Constant Derivative Function

Let \(g:\mathbb{R}\to\mathbb{R}\) be \(g(x)=4x-3\). Find \(g'\) using the difference quotient.

For an arbitrary fixed \(x\), first compute \(g(x+h)\):

$$ g(x+h)=4(x+h)-3=4x+4h-3. $$

Then form and simplify the quotient:

$$ \begin{aligned} \frac{g(x+h)-g(x)}{h} &=\frac{(4x+4h-3)-(4x-3)}{h}\\ &=\frac{4x+4h-3-4x+3}{h}\\ &=\frac{4h}{h}\\ &=4. \end{aligned} $$

Taking the limit gives

$$ g'(x)=\lim_{h\to 0}4=4. $$

This holds for every \(x\in\mathbb{R}\), so \(g'\) is the constant function \(g':\mathbb{R}\to\mathbb{R}\) with \(g'(x)=4\). To check, substituting the result into the difference quotient gives the same value \(4\) for every nonzero \(h\), so its limit is indeed \(4\) at every input.

This example shows that the derivative function can happen to be constant. That is a particular outcome, not the meaning of derivative function in general. In the square-function example, the derivative depends on \(x\); here it does not.

Why the Derivative Formula Depends on the Input

The derivative is defined point by point. At each fixed \(x\), we examine the difference quotient near that input and take its limit. Repeating that process for all eligible inputs gives the derivative function. When the resulting limits vary with \(x\), the derivative function records that variation.

1
Choose an arbitrary input.
Keep \(x\) symbolic so the calculation can apply across the domain, not just at one selected point.
2
Form the difference quotient.
Compare \(f(x+h)\) with \(f(x)\), divide by the nonzero input change \(h\), and simplify.
3
Take the limit in \(h\).
If a finite limit exists, its value is the derivative at that fixed \(x\).
4
Collect the values across inputs.
The rule obtained as \(x\) varies is the derivative function on the set where those limits exist.

The method works because the definition specifies exactly how to find the derivative at each input. Keeping \(x\) arbitrary allows one calculation to describe the result at every input for which the limit exists. If one substitutes a number for \(x\) at the beginning, the calculation instead addresses only that one input.

Worked Example: A Cubic Produces a Quadratic Derivative Function

Worked Example: A Cubic Produces a Quadratic Derivative Function

Let \(p:\mathbb{R}\to\mathbb{R}\) be \(p(x)=x^3-2x\). Find its derivative function directly from the definition.

For fixed \(x\), evaluate \(p(x+h)\):

$$ p(x+h)=(x+h)^3-2(x+h) =x^3+3x^2h+3xh^2+h^3-2x-2h. $$

Subtract \(p(x)=x^3-2x\) and divide by \(h\ne 0\):

$$ \begin{aligned} \frac{p(x+h)-p(x)}{h} &=\frac{x^3+3x^2h+3xh^2+h^3-2x-2h-(x^3-2x)}{h}\\ &=\frac{3x^2h+3xh^2+h^3-2h}{h}\\ &=3x^2+3xh+h^2-2. \end{aligned} $$

Now let \(h\to 0\):

$$ p'(x)=\lim_{h\to 0}(3x^2+3xh+h^2-2)=3x^2-2. $$

Thus \(p':\mathbb{R}\to\mathbb{R}\) is given by \(p'(x)=3x^2-2\). To check the result, subtract this proposed limit from the simplified quotient:

$$ (3x^2+3xh+h^2-2)-(3x^2-2)=3xh+h^2. $$

For fixed \(x\), \(3xh+h^2\to 0\) as \(h\to 0\). Therefore the quotient approaches \(3x^2-2\), as claimed.

A Short Proof That the Derivative Assignment Is Unambiguous

At any fixed input, the difference quotient has at most one limit. This guarantees that the derivative definition assigns no more than one output to each input.

Proof: Fix an input \(x\), and write \(Q(h)=\frac{f(x+h)-f(x)}{h}\) for nonzero \(h\) close to zero. Suppose \(Q(h)\) tends both to \(L\) and to \(M\). For every \(\varepsilon>0\), the definition of these two limits gives a sufficiently small nonzero \(h\) for which \(|Q(h)-L|<\varepsilon/2\) and \(|Q(h)-M|<\varepsilon/2\). The triangle inequality then gives $$ |L-M|\le |L-Q(h)|+|Q(h)-M|<\varepsilon. $$ Since this holds for every \(\varepsilon>0\), \(L=M\). Thus, wherever the derivative limit exists, it has exactly one value. Assigning that unique value to each input in \(D\) defines the function \(f':D\to\mathbb{R}\).

A Common Mistake: Fixing the Input Too Soon

Suppose \(f(x)=x^2\), and the goal is to find the derivative function. A common mistake is to choose \(x=2\) before completing the calculation, then conclude that the derivative is \(4\). That calculation can find \(f'(2)\), but it cannot establish the derivative function at every input.

$$ \frac{(2+h)^2-2^2}{h} =\frac{4+4h+h^2-4}{h} =4+h \quad\Longrightarrow\quad f'(2)=4. $$

The correction is to keep \(x\) arbitrary until after taking the limit:

$$ \frac{(x+h)^2-x^2}{h}=2x+h \quad\Longrightarrow\quad f'(x)=2x. $$

Only after obtaining the full rule should a particular input be substituted. Another related error is to assume that \(f'\) must have exactly the same domain as \(f\). The derivative function is defined only at inputs where the derivative limit exists.

Why the Derivative Function Matters

Thinking of \(f'\) as a function makes it possible to describe how rates vary across inputs. A calculation at one input answers a local question; the derivative function gathers the answers into one rule. That rule can depend on its input, as \(2x\) and \(3x^2-2\) do, or it can be constant, as \(4\) is.

When working with derivatives, first be clear about which object is being requested. If the task asks for a derivative function, preserve the variable \(x\) and state the domain where the derivative exists. If it asks for the derivative at a specified input \(a\), evaluate the derivative there to obtain the single number \(f'(a)\). The distinction prevents a pointwise calculation from being mistaken for a general rule.

Check Your Understanding

Answer these questions using the definitions and examples above.

  1. In the definition of \(f'(x)\), which quantity is fixed while the limit is taken, and which quantity approaches zero?
  2. For \(f(x)=x^2\), explain the difference between the statements \(f'(x)=2x\) and \(f'(3)=6\).
  3. Can a derivative function be constant? Give an example from this tutorial and state its derivative function.
  4. What determines the domain of \(f'\)?
  5. Why does computing the difference quotient at one numerical input not, by itself, find the derivative function?

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