Using Differentials to Approximate Finite Changes

From Instantaneous Rate to Finite Change

A derivative describes an instantaneous rate of change at a particular input. In applications, however, we often want to estimate what happens when an input changes by a small but nonzero amount. For example, if a radius increases slightly, how much does the circle’s area increase? The derivative gives a practical estimate without requiring us to compute the new function value exactly.

The key is to keep two quantities distinct. The change in the function is the actual finite change. The differential uses the derivative at the starting point to predict that change. These quantities are related, but they are not generally equal.

Definition: Let $y=f(x)$, let $x$ be a base input, and let the input change by $\Delta x$. The actual finite change in the output is $$ \Delta y=f(x+\Delta x)-f(x). $$ The differential is defined by $$ dy=f'(x)\,dx. $$ To use the differential to estimate the change caused by $\Delta x$, set $dx=\Delta x$. Then $dy=f'(x)\Delta x$ is the linear estimate for $\Delta y$.

The function must be differentiable at the base point $x$ for this formula to use a finite derivative $f'(x)$. The step $\Delta x$ must also be chosen so that $f(x+\Delta x)$ is defined if we want to calculate the actual change for comparison. A small step often makes a useful local estimate, but the differential formula itself does not claim that $dy$ equals $\Delta y$.

Why the Differential Estimates the Change

The derivative at $x$ is the limit of the average rate of change as the input step approaches zero. For a nonzero step $h$, the average rate is $\frac{f(x+h)-f(x)}{h}$. If $f$ is differentiable at $x$, this average rate approaches $f'(x)$ as $h$ approaches zero.

We can make the relationship with the differential precise. Let $h=\Delta x$, and define the difference between the average rate and the derivative by $\varepsilon(h)=\frac{f(x+h)-f(x)}{h}-f'(x)$, for nonzero $h$. Differentiability says that $\varepsilon(h)$ approaches zero as $h$ approaches zero. Multiplying the defining equation by $h$ gives the actual change as the differential plus a remaining term:

\begin{aligned} f(x+h)-f(x) &=h\left(f'(x)+\varepsilon(h)\right)\\ &=f'(x)h+h\varepsilon(h). \end{aligned}

When $dx=h$, the first term $f'(x)h$ is $dy$. The remaining term $h\varepsilon(h)$, divided by $h$, approaches zero as the step shrinks. Thus the actual change and the differential become close relative to the size of the input step. This is why the derivative at the base point supplies a local linear estimate.

Takeaway: For a small finite input change, estimate the output change using the derivative at the starting input: $$ \Delta y \approx dy=f'(x)\Delta x. $$ The approximation symbol matters: $dy$ is a prediction, while $\Delta y$ is the actual change.

Worked Examples

Worked Example: Estimating a Change in a Square

Let $f(x)=x^2$. Estimate the change in $f$ when $x$ increases from $4$ to $4.1$, then compare the estimate with the actual change.

The base input is $x=4$, and the increment is $\Delta x=4.1-4=0.1$. Differentiate $f$:

f'(x)=2x,\qquad f'(4)=8.

Set $dx=\Delta x=0.1$. The differential estimate is

dy=f'(4)\,dx=8(0.1)=0.8.

Now calculate the actual change:

\Delta y=f(4.1)-f(4)=(4.1)^2-4^2=16.81-16=0.81.

Thus the differential predicts an increase of $0.8$, while the actual increase is $0.81$. The comparison can be checked directly by expanding the square: $(4+0.1)^2-4^2=16+0.8+0.01-16=0.81$. The extra $0.01$ is why the estimate is not an exact change.

Worked Example: Estimating a Change in a Square Root

Estimate the change in $f(x)=\sqrt{x}$ when $x$ increases from $25$ to $26$, and compare the estimate with the actual change.

Here $x=25$ and $\Delta x=1$. The derivative and its value at the base point are

f'(x)=\frac{1}{2\sqrt{x}}, \qquad f'(25)=\frac{1}{2(5)}=\frac{1}{10}.

Using $dx=1$, the differential estimate is

dy=f'(25)\,dx=\frac{1}{10}(1)=0.1.

The actual change is

\Delta y=\sqrt{26}-\sqrt{25}=\sqrt{26}-5.

To evaluate it, $\sqrt{26}\approx5.0990195$, so $\Delta y\approx0.0990195$. The differential predicts an increase of $0.1$. As a check on the square-root value, $5.0990195^2\approx26.000000$, to the displayed precision; subtracting $5$ gives approximately $0.0990195$. The estimated change is close to, but not identical to, the actual change.

Worked Example: A Change in Circle Area

A circle has radius $r=10$ centimeters. Estimate the change in its area if the radius increases by $0.2$ centimeter, then compare the estimate with the actual change. The area as a function of radius is $A(r)=\pi r^2$.

Differentiate with respect to $r$:

A'(r)=2\pi r,\qquad A'(10)=20\pi.

The radius increment is $dr=\Delta r=0.2$ centimeters. Therefore, the differential estimate is

dA=A'(10)\,dr=(20\pi)(0.2)=4\pi\ \text{cm}^2 \approx12.566371\ \text{cm}^2.

The new radius is $10.2$ centimeters, so the actual change is

\begin{aligned} \Delta A &=A(10.2)-A(10)\\ &=\pi(10.2)^2-\pi(10)^2\\ &=\pi(104.04-100)\\ &=4.04\pi\ \text{cm}^2\\ &\approx12.691504\ \text{cm}^2. \end{aligned}

The differential estimate is $4\pi\ \text{cm}^2$, and the actual increase is $4.04\pi\ \text{cm}^2$. To check the expansion, $(10+0.2)^2=100+4+0.04=104.04$, so the area difference is indeed $4.04\pi\ \text{cm}^2$. The units also agree: $A'(r)$ has units of square centimeters per centimeter, and multiplying by a change in centimeters gives square centimeters.

Worked Example: A Negative Input Change

Let $f(x)=1/x$. Estimate the change in $f$ when $x$ decreases from $2$ to $1.9$, and compare with the actual change.

The base input is $x=2$, and the signed increment is $\Delta x=1.9-2=-0.1$. Since $f'(x)=-1/x^2$,

f'(2)=-\frac{1}{2^2}=-\frac14.

Use $dx=-0.1$, including its negative sign:

dy=f'(2)\,dx=-\frac14(-0.1)=0.025.

The actual change is

\begin{aligned} \Delta y &=f(1.9)-f(2)\\ &=\frac{1}{1.9}-\frac12\\ &=\frac{10}{19}-\frac12\\ &=\frac{20-19}{38}\\ &=\frac{1}{38}\\ &\approx0.026316. \end{aligned}

The input decreased, but the output increased: the negative derivative and negative input increment produce a positive differential. The exact value $1/38$ is positive as well, and $1/38\approx0.026316$, compared with the estimate $0.025$.

Keeping the Quantities Straight

Both $\Delta x$ and $dx$ describe an input step in this application: set $dx=\Delta x$. But the output quantities have different roles. The symbol $\Delta y$ means the actual difference between the function’s endpoint values. The symbol $dy$ means the derivative at the base point multiplied by the selected input step.

The estimate uses the derivative at the starting input $x$, not at the endpoint $x+\Delta x$. The sign of $\Delta x$ must also be preserved. A negative input step multiplied by a negative derivative gives a positive estimate, as in the reciprocal example.

Units provide another useful check. If $y=f(x)$ is measured in meters and $x$ in seconds, then $f'(x)$ has units of meters per second. Multiplying by $dx$, measured in seconds, gives $dy$ in meters. A result with the wrong units usually signals an incorrect derivative or an incorrect interpretation of the increment.

A Common Mistake

A common error is to state that the differential is exactly the finite change. For $f(x)=x^2$, starting at $x=4$ with $\Delta x=0.1$, the differential is $dy=8(0.1)=0.8$. Claiming that the actual change is therefore exactly $0.8$ is incorrect. The endpoint calculation gives $\Delta y=(4.1)^2-4^2=16.81-16=0.81$.

The correction is to say that $dy=0.8$ is the linear estimate for $\Delta y=0.81$. The derivative supplies the local rate at the base point; it does not, by itself, calculate every contribution to a nonzero finite change.

When Differentials Are Useful

Differentials are useful when a direct evaluation at the new input is inconvenient, when a quick estimate is sufficient, or when the relationship between an input adjustment and an output adjustment is more important than an exact value. The same process applies to formulas from geometry, physical models, and other settings: identify the function, choose the base input, find its derivative there, and multiply by the signed input increment.

For any particular step, the actual change can be compared with the differential when it can be computed. Such a comparison shows how close that estimate was for that case; it does not establish a general guarantee for other inputs or step sizes. The central skill is therefore not merely calculating $f'(x)\Delta x$, but identifying it correctly as an estimate of the finite change.

Check Your Understanding

For each question, distinguish the differential estimate from the actual finite change.

If $f(x)=x^2$, what differential estimates the change from $x=3$ to $x=3.2$?
For $f(x)=\sqrt{x}$, identify the base input and signed increment when estimating the change from $x=16$ to $x=15.5$.
In $dy=f'(x)\,dx$, what choice of $dx$ estimates the change caused by $\Delta x$?
Why does differentiability at the base point make $dy$ a useful local estimate?
A quantity has units of liters and its input has units of minutes. What units should its differential have when it estimates a change in the quantity?

Using Differentials to Approximate Finite Changes

What You'll Learn

From Instantaneous Rate to Finite Change

Why the Differential Estimates the Change

Worked Examples

Worked Example: Estimating a Change in a Square