From an Input Change to a Predicted Output Change
A derivative describes the slope of a function at a point. That slope can also be used to answer a practical question: if the input changes by a chosen amount, what output change does the tangent line predict?
Differentials give a concise notation for this prediction. The input change is written \(dx\), and the corresponding tangent-line prediction for the output change is written \(dy\). The calculation uses the derivative at a specified base point, so it is important to identify that point before using the notation.
The derivative \(f'(x)\) is the tangent-line slope at the current input \(x\). Multiplying this slope by the chosen horizontal change \(dx\) gives the vertical change along that tangent line. Thus a differential is not a separate way to calculate a derivative: it uses the derivative to predict an output change.
Deriving the Tangent-Line Interpretation
For a differentiable function \(f\), the tangent line at the point \((x,f(x))\) has slope \(f'(x)\). If \(u\) denotes a variable input on that line, its equation is
Now move along the tangent line from input \(x\) to input \(x+dx\). Substituting \(u=x+dx\) into the tangent-line equation gives the predicted output at the new input. Subtracting the original output \(f(x)\) gives the predicted change:
This establishes the interpretation directly: \(dy\) is exactly the change in the tangent line’s output as its input moves by \(dx\). The result depends on using the derivative at the chosen base point \(x\). If the base point changes, the slope—and therefore the differential for the same \(dx\)—may change too.
Signs and Units
The sign of \(dy\) follows from the signs of \(f'(x)\) and \(dx\). If the slope and input change have the same sign, the predicted output change is positive. If they have opposite signs, the prediction is negative. A zero input change gives \(dy=0\).
Units provide another useful check. If \(y\) is measured in metres and \(x\) in seconds, then \(f'(x)\) has units metres per second. Multiplying by \(dx\), measured in seconds, gives \(dy\) in metres:
In general, the units of \(dy\) are the units of \(f'(x)\) multiplied by the units of \(dx\), which are the units of the output \(y\). If the units do not match, check whether the derivative, input change, or base point has been interpreted incorrectly.
Worked Examples
Worked Example: A Quadratic at a Given Base Point
Let \(f(x)=x^2\). At the base point \(x=3\), the derivative is \(f'(3)=6\). Find the differential \(dy\) for the input change \(dx=0.1\), and verify it using the tangent line.
Use the differential definition, evaluating the slope at the base point:
To check the tangent-line interpretation, the tangent line at \(x=3\) passes through \((3,9)\) and has slope \(6\). At the shifted input \(3+0.1=3.1\), its predicted output is
The tangent-line prediction of the output change is \(9.6-9=0.6\), which agrees with \(dy=0.6\). The differential is the tangent-line change for the specified input increment.
Worked Example: A Negative Slope and a Negative Input Change
Suppose a differentiable function \(g\) satisfies \(g(2)=5\) and \(g'(2)=-4\). For the input change \(dx=-0.25\), find \(dy\) and check the result using the tangent line.
The base point is \(2\), so use the slope \(g'(2)=-4\):
The shifted input is \(2+(-0.25)=1.75\). The tangent line at the base point has equation \(y_{\text{tan}}=5-4(u-2)\). At \(u=1.75\), its output is
The tangent-line output has changed from \(5\) to \(6\), a predicted change of \(6-5=1\), as given by \(dy\). Both \(dx\) and the slope are negative, so their product is positive. Moving left along a line with negative slope raises the line’s output.
Worked Example: Predicting an Area Change with Units
The area of a circle is \(A(r)=\pi r^2\), where \(r\) is measured in centimetres and \(A\) in square centimetres. At radius \(r=5\) cm, the derivative is \(A'(5)=10\pi\) cm. Find the differential area \(dA\) for an input change \(dr=0.02\) cm, and verify it using the tangent line.
Use the derivative at the stated base radius and multiply by the radius change:
The units multiply to square centimetres, as required for an area change. Numerically, \(0.2\pi\approx0.628\), so the prediction is approximately \(0.628\ \text{cm}^2\), rounded to three decimal places.
The tangent line to \(A(r)\) at \(r=5\) is \(A_{\text{tan}}=25\pi+10\pi(r-5)\). At the shifted radius \(5+0.02=5.02\) cm, this line predicts
The tangent-line prediction of the area change is therefore \((25\pi+0.2\pi)-25\pi=0.2\pi\ \text{cm}^2\), matching \(dA\). This check confirms both the value and the units.
Choosing and Using the Base Point
A differential calculation is tied to a particular point. The value \(f'(x)\) in \(dy=f'(x)\,dx\) is the slope at the base input \(x\), while \(dx\) specifies how far the input moves from that base. Keeping those roles separate prevents a common substitution error.
Record the input \(x\) at which the tangent slope is known or evaluated.
Use \(dx\) for the chosen increment from that base point. The shifted input is \(x+dx\).
Calculate \(dy=f'(x)\,dx\), retaining signs and units.
Read \(dy\) as the tangent line’s predicted output change for that input increment.
A frequent mistake is to use the slope at the shifted input instead. For example, in the quadratic example, using the slope at \(3.1\) gives \(f'(3.1)=6.2\), and then \(6.2(0.1)=0.62\). That is not the differential based at \(x=3\): it uses a different tangent slope. The correct calculation is \(f'(3)(0.1)=6(0.1)=0.6\), because the base point is \(3\). The tangent-line check at that base point also gives a predicted change of \(0.6\).
What a Differential Does—and Does Not—Say
The definition assigns a tangent-line prediction to an input increment. It does not, by itself, assert that the function’s own output changes by exactly that amount. The distinction matters: \(dy\) is defined from the slope at the base point and the chosen \(dx\), while the function’s value at a shifted input is a separate quantity.
For a straight-line function, the tangent line agrees with the function, so the tangent prediction gives the exact change. For a curved function, the tangent line is a local model. The differential still has a precise meaning as the change along that line; it should not be mistaken for a statement that the function itself is linear.
This notation is useful because it translates a derivative, which describes a rate or slope, into an output change associated with a particular input change. It also keeps dimensions visible: rates multiplied by input increments produce output increments. In applications, that makes it easier to express a local prediction in the same units as the quantity being studied.
Check Your Understanding
Use the base point specified in each question, and interpret each differential as a tangent-line prediction.
- If \(f'(4)=3\) and \(dx=-0.2\), what is \(dy\), and what do its sign and value mean?
- A tangent line has slope \(-5\) at a base point. What differential does it predict for \(dx=0.4\)?
- For \(y=f(x)\), explain why the differential uses \(f'(x)\) rather than the slope at \(x+dx\).
- If \(x\) is measured in seconds and \(y\) in metres, what units must \(f'(x)\) and \(dy\) have?
- In one sentence, explain why \(dy\) is a tangent-line prediction rather than automatically the exact output change of the function.
