From a Tangent Line to Differentials
A derivative describes the slope of a function at a particular input. A tangent line uses that slope to describe how the function’s output is expected to respond locally when the input changes. Differentials give names to the input change and the tangent line’s predicted output change.
For a function \(y=f(x)\), the notation \(dx\) represents a chosen change in the input. The notation \(dy\) represents the change in output predicted by the tangent line at the current input. These quantities are related by \(dy=f'(x)dx\). Understanding this relationship requires keeping the tangent-line prediction distinct from the function’s actual change.
Definition of \(dx\) and \(dy\)
Fix an input \(x\) at which \(f\) is differentiable. Choose an input change \(dx\). The tangent line at \(x\) has slope \(f'(x)\), so its output changes by its slope multiplied by the horizontal change \(dx\).
In this definition, \(dx\) is the input change supplied to the tangent-line model. Once \(x\) and \(dx\) are specified, \(dy\) is determined by the derivative at \(x\). For fixed \(x\), \(dy\) depends linearly on \(dx\): doubling \(dx\) doubles \(dy\), and reversing the sign of \(dx\) reverses the sign of \(dy\).
The relationship does not require \(dx\) to be an infinitesimal number. In AP Calculus, it is useful to regard \(dx\) as an ordinary real input change and \(dy\) as the corresponding tangent-line change. The interpretation is local: the tangent line captures the function’s behavior near the point, and is especially informative for small input changes.
Deriving the Differential Relationship
The derivation follows directly from the linearization formula. At a fixed base point \(a\), the linearization of \(f\) is
Here \(t\) is an input near \(a\), and \(L_a(t)\) is the output given by the tangent-line model. The predicted output change is the tangent-line value at the new input minus its value at the base input. Writing the input change as \(dx\) gives the differential.
The tangent line agrees with the function at \(a\), so \(L_a(a)=f(a)+f'(a)(a-a)=f(a)\).
The new input in the linear model is \(a+dx\), giving \(L_a(a+dx)=f(a)+f'(a)((a+dx)-a)=f(a)+f'(a)dx\).
The tangent-line output change is \(L_a(a+dx)-L_a(a)=f'(a)dx\). By definition, this predicted change is \(dy\).
Therefore the differential relationship is \(dy=f'(a)dx\) at the base point \(a\). Since the base point can be denoted by \(x\), the standard notation is \(dy=f'(x)dx\). The derivative supplies the tangent-line slope; multiplying by the horizontal change gives the tangent line’s vertical change.
Differential and Actual Change
The function’s actual output change for an input change \(dx\) is often denoted by \(\Delta y\). If the starting input is \(x\), then the new input is \(x+dx\), and
This expression uses the function itself. In contrast, \(dy=f'(x)dx\) uses the tangent line at \(x\). They describe related but different quantities: \(\Delta y\) is the actual change, while \(dy\) is the linear model’s predicted change.
| Symbol | Meaning | How it is found |
|---|---|---|
| \(dx\) | Chosen input change | Specified for the situation |
| \(\Delta y\) | Actual change in function output | \(f(x+dx)-f(x)\) |
| \(dy\) | Tangent-line predicted output change | \(f'(x)dx\) |
The distinction matters even when \(dy\) and \(\Delta y\) are close. The equality \(dy=\Delta y\) is not part of the definition. A differential describes the linearization; it does not redefine the function’s exact change.
Worked Examples
Worked Example: A Polynomial at a Chosen Input
Let \(y=x^3-2x\). Find the differential \(dy\) at \(x=2\) when \(dx=0.1\), and verify it using the tangent line.
First differentiate:
At \(x=2\), the slope is
Now use the definition of the differential:
To check, \(f(2)=2^3-2(2)=8-4=4\), so the tangent line at \(x=2\) is \(L_2(t)=4+10(t-2)\). At the input \(2+0.1=2.1\),
The tangent-line output change is \(L_2(2.1)-L_2(2)=5-4=1\), matching \(dy=1\). This check verifies the differential as a change in the linear model.
Worked Example: A Square-Root Function with a Negative Input Change
Let \(y=\sqrt{x}\). Find \(dy\) at \(x=4\) when \(dx=-0.16\), and check the result using the tangent line.
Differentiate:
At the base point \(x=4\),
Therefore,
The tangent line at \(x=4\), where \(f(4)=2\), is \(L_4(t)=2+\frac14(t-4)\). The changed input is \(4+(-0.16)=3.84\). Substitution gives
At the base point, \(L_4(4)=2\). Thus the tangent-line change is \(1.96-2=-0.04\), which agrees with the computed differential. The negative sign is consistent with a negative input change and a positive slope.
Worked Example: A Trigonometric Function in Radians
Let \(y=\sin x\), with angles measured in radians. Find \(dy\) at \(x=\frac{\pi}{6}\) for \(dx=0.02\), and check using the tangent line.
The derivative is \(f'(x)=\cos x\). At the base point,
The differential is therefore
Since \(\sin(\frac{\pi}{6})=\frac12\), the tangent line is \(L_{\pi/6}(t)=\frac12+\frac{\sqrt{3}}{2}(t-\frac{\pi}{6})\). At the changed input \(t=\frac{\pi}{6}+0.02\), its value is
Subtracting the starting tangent-line value \(\frac12\) gives \(\frac{\sqrt{3}}{100}\), exactly the differential found above.
Units and Interpretation
The differential relationship also keeps units consistent. If \(x\) is measured in meters and \(y\) in square meters, then \(f'(x)\) has units of square meters per meter. Multiplying by \(dx\), measured in meters, gives \(dy\) in square meters:
This is another way to interpret the derivative: it measures output change per unit of input change. A differential is the tangent-line output change associated with an input change, so its units must match the units of the output.
A Common Mistake
A frequent error is to calculate the derivative and report that value as \(dy\), without multiplying by the input change. For \(f(x)=x^2\) at \(x=3\), the slope is \(f'(3)=2(3)=6\). If \(dx=0.2\), the incorrect calculation is
That number is the tangent-line slope, not the tangent-line output change. The input change is \(0.2\), so the correct calculation is
The correction is to keep the roles separate: \(f'(x)\) gives change in output per unit input, while \(dy\) gives the predicted output change for the specified \(dx\). Similarly, do not automatically replace the actual change \(\Delta y\) by \(dy\); the former comes from the function, and the latter from its tangent line.
Check Your Understanding
Answer each question using the distinction between an input change, an actual output change, and a tangent-line output change.
- For \(y=f(x)\), what does \(dx\) represent, and how is \(dy\) defined?
- If \(f'(a)=-4\) and \(dx=0.3\), what is \(dy\) at \(x=a\)?
- Write a formula for the actual output change \(\Delta y\) when the input moves from \(x\) to \(x+dx\).
- In the linearization \(L_a(t)=f(a)+f'(a)(t-a)\), show how subtracting \(L_a(a)\) from \(L_a(a+dx)\) gives the differential formula.
- Why is \(dy\) not automatically equal to the function’s actual change \(\Delta y\)?
