Two Different Changes
A function changes when its input changes. Calculus gives us two useful ways to describe that change: one records the exact difference between two function values, and the other uses the derivative at the starting input to describe the change along the tangent line.
These quantities are related, but they are not generally equal. Keeping their definitions distinct is important whenever a differential is used to represent a change.
The actual change \(\Delta y\) is defined by subtracting the two function values. The differential \(dy\), by contrast, is computed from the derivative at the starting point and the chosen input increment. It is the change on the tangent line over that horizontal step.
The Tangent-Line Interpretation
At the point \((x,f(x))\), the tangent line has slope \(f'(x)\). Moving horizontally by \(dx\) along this line changes its vertical coordinate by \(f'(x)\,dx\), which is exactly \(dy\). The graph of the function need not follow the tangent line over a finite step, so its actual vertical change may differ.
For a direct geometric comparison, let \(P=(x,f(x))\) and \(Q=(x+\Delta x,f(x+\Delta x))\) be points on the graph. The vertical difference between them is \(\Delta y\). At the same horizontal position \(x+\Delta x\), the tangent line through \(P\) has vertical change \(dy=f'(x)\Delta x\). Thus \(\Delta y\) is the change on the graph, while \(dy\) is the change on the tangent line.
The secant line through \(P\) and \(Q\) has slope \(\Delta y/\Delta x\), provided \(\Delta x\ne 0\). The tangent line at \(P\) has slope \(f'(x)\). The two slopes describe different lines: one joins two points on the graph, and the other describes the graph’s instantaneous slope at the starting point.
Worked Examples
Worked Example: A Quadratic Function
Let \(f(x)=x^2\), start at \(x=2\), and take \(\Delta x=dx=\frac{1}{10}\). Calculate the actual change and the differential.
First calculate the actual change. The new input is $$ x+\Delta x=2+\frac{1}{10}=\frac{21}{10}. $$ Therefore, $$ f\left(\frac{21}{10}\right)=\left(\frac{21}{10}\right)^2=\frac{441}{100}, \qquad f(2)=2^2=4=\frac{400}{100}. $$ So $$ \Delta y=f\left(2+\frac{1}{10}\right)-f(2) =\frac{441}{100}-\frac{400}{100} =\frac{41}{100}. $$
Now \(f'(x)=2x\), so \(f'(2)=4\). The differential is $$ dy=f'(2)\,dx=4\left(\frac{1}{10}\right)=\frac{4}{10}=\frac{2}{5}=\frac{40}{100}. $$ Thus \(\Delta y=\frac{41}{100}\), whereas \(dy=\frac{40}{100}\); they differ by \(\frac{1}{100}\).
As a geometric check, the secant slope is $$ \frac{\Delta y}{\Delta x} =\frac{41/100}{1/10} =\frac{41}{10}, $$ while the tangent slope at \(x=2\) is \(f'(2)=4\). These are different slopes, as expected for the graph over this finite step.
Worked Example: A Cubic Function
Let \(f(x)=x^3\), start at \(x=1\), and take \(\Delta x=dx=\frac{1}{10}\).
The new input is \(1+\frac{1}{10}=\frac{11}{10}\), so $$ f\left(\frac{11}{10}\right)=\left(\frac{11}{10}\right)^3=\frac{1331}{1000}, \qquad f(1)=1=\frac{1000}{1000}. $$ It follows that the actual change is $$ \Delta y=\frac{1331}{1000}-\frac{1000}{1000} =\frac{331}{1000}. $$
The derivative is \(f'(x)=3x^2\), and hence \(f'(1)=3\). The differential is $$ dy=3\left(\frac{1}{10}\right) =\frac{3}{10} =\frac{300}{1000}. $$ Therefore, the two changes are \(\frac{331}{1000}\) and \(\frac{300}{1000}\), respectively.
For a geometric check, the secant slope is $$ \frac{\Delta y}{\Delta x} =\frac{331/1000}{1/10} =\frac{331}{100} =3.31. $$ The tangent slope at the starting input is \(f'(1)=3\). The secant slope is the average change across this step; the tangent slope is the slope at \(x=1\).
Worked Example: A Reciprocal Function
Let \(f(x)=\frac{1}{x}\), start at \(x=2\), and take \(\Delta x=dx=-\frac{1}{5}\). The input moves to the left, since the increment is negative.
The new input is $$ x+\Delta x=2-\frac{1}{5}=\frac{9}{5}. $$ The function values are $$ f\left(\frac{9}{5}\right)=\frac{5}{9}, \qquad f(2)=\frac{1}{2}. $$ Thus the actual change is $$ \Delta y=\frac{5}{9}-\frac{1}{2} =\frac{10}{18}-\frac{9}{18} =\frac{1}{18}. $$
Since \(f'(x)=-\frac{1}{x^2}\), we have \(f'(2)=-\frac{1}{4}\). Therefore, $$ dy=f'(2)\,dx =-\frac{1}{4}\left(-\frac{1}{5}\right) =\frac{1}{20}. $$ Both changes are positive: decreasing the input from \(2\) to \(\frac{9}{5}\) increases the value of the reciprocal function. The actual change is \(\frac{1}{18}\); the differential is \(\frac{1}{20}\).
The secant slope is $$ \frac{\Delta y}{\Delta x} =\frac{1/18}{-1/5} =-\frac{5}{18}. $$ The tangent slope at \(x=2\) is \(-\frac{1}{4}\). Both slopes are negative, although the vertical changes are positive, because the horizontal increment is negative.
Why the Differential Is a Tangent-Line Change
The connection follows directly from the tangent-line formula. The tangent line to the graph of \(f\) at \(x\) has equation $$ L(t)=f(x)+f'(x)(t-x). $$ If the horizontal input changes from \(x\) to \(x+\Delta x\), the tangent-line value changes by $$ L(x+\Delta x)-L(x) =f(x)+f'(x)\Delta x-f(x) =f'(x)\Delta x. $$ When \(dx=\Delta x\), this difference is precisely \(dy\).
The function’s actual change is instead \(f(x+\Delta x)-f(x)\). The two definitions can be compared exactly by adding and subtracting the tangent-line change: $$ \Delta y =f'(x)\Delta x+ \left(f(x+\Delta x)-f(x)-f'(x)\Delta x\right). $$ The first term is \(dy\) when \(dx=\Delta x\). The expression in parentheses records the difference between the actual change and the tangent-line change. This identity does not say that the difference is zero; the examples show that it need not be.
Differentiability explains why the tangent line is useful for describing changes close to the starting input. By the definition of the derivative, $$ \lim_{\Delta x\to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x} =f'(x). $$ For nonzero \(\Delta x\), multiplying the quotient by \(\Delta x\) gives $$ \Delta y =\left(\frac{f(x+\Delta x)-f(x)}{\Delta x}\right)\Delta x. $$ As \(\Delta x\) approaches zero, the factor in parentheses approaches \(f'(x)\). This is why \(f'(x)\Delta x\), the differential when \(dx=\Delta x\), describes the function’s change locally. It does not make the two quantities identical for every finite step.
When Are the Changes Equal?
For an affine function \(f(x)=mx+b\), the differential and actual change agree for every permitted input increment when \(dx=\Delta x\). Indeed, $$ \Delta y=f(x+\Delta x)-f(x) =\bigl(m(x+\Delta x)+b\bigr)-(mx+b) =m\Delta x, $$ and $$ dy=f'(x)\,dx=m\Delta x. $$ For a nonlinear function, the two quantities are generally different, although they can coincide for particular starting points or increments. The definitions alone do not justify assuming equality.
Common Pitfall: Replacing the Actual Change by the Differential
A common mistake is to calculate \(dy\) and then label it \(\Delta y\), as if the two symbols named the same quantity. For example, take \(f(x)=x^2\), \(x=3\), and \(\Delta x=dx=\frac{1}{5}\). A mistaken calculation might say $$ \Delta y=dy=f'(3)\left(\frac{1}{5}\right) =6\left(\frac{1}{5}\right) =\frac{6}{5}. $$ The computation on the right gives the differential, not necessarily the actual change.
To find the actual change, evaluate both function values: $$ f\left(3+\frac{1}{5}\right) =f\left(\frac{16}{5}\right) =\frac{256}{25}, \qquad f(3)=9=\frac{225}{25}. $$ Therefore, $$ \Delta y=\frac{256}{25}-\frac{225}{25} =\frac{31}{25}. $$ Meanwhile, $$ dy=f'(3)\,dx=6\left(\frac{1}{5}\right)=\frac{6}{5}=\frac{30}{25}. $$ The corrected conclusion is \(\Delta y=\frac{31}{25}\) and \(dy=\frac{30}{25}\). The derivative \(f'(3)=6\) is the tangent slope; multiplying it by the input increment gives the differential.
Check Your Understanding
For each question, keep the exact function change separate from the tangent-line change.
- For a differentiable function, write the definitions of \(\Delta y\) and \(dy\). What must be set equal to compare them over the same input step?
- For \(f(x)=x^2\), \(x=1\), and \(\Delta x=dx=\frac{1}{4}\), calculate \(\Delta y\) and \(dy\) exactly.
- Geometrically, which change is measured by the secant between two graph points, and which is measured along the tangent at the starting point?
- For what class of functions do \(dy\) and \(\Delta y\) agree for every permitted increment when \(dx=\Delta x\)?
- Why does the derivative definition explain the usefulness of \(dy\) for small input increments without asserting equality for every finite increment?
