Calculus I · Unit 2B · lesson

Differentials

Concept

Learning objectives

Use $dy=f'(x)dx$ to estimate changes in an output.

Differentials and Small Changes

Explanation

Before the formulas

The approximation in Differentials replaces a nonlinear function with its best local line. The known value supplies the starting height, and the derivative supplies the rate of change. The estimate is useful because lines are easy to calculate with, not because the original curve has become exactly linear.

Always identify the center $a$ and the target $x$ . Write $\Delta x=x-a$ , calculate the predicted change $f'(a)\Delta x$ , and then add it to $f(a)$ . Use concavity or a numerical comparison to understand the direction and size of the error.

**Visual study**
The curve produces the actual output change y. The tangent line produces the differential dy=f'(x)dx. Their vertical gap is the approximation error.

Explanation

Differentials package the tangent-line prediction

For an input change $dx$ , the differential $dy=f'(x)dx$ is the change predicted by the tangent line. The actual change is $\Delta y=f(x+dx)-f(x)$ . Near the base point, $dy$ and $\Delta y$ are close but not generally equal.

This notation is especially useful in measurement problems because it keeps the estimated output error attached to the input uncertainty and local sensitivity.

Differentials package the linear approximation into the compact relation

dy=f'(x)\,dx.

Here $dx$ represents a chosen small input change, and $dy$ represents the corresponding linearized output change. The actual change $\Delta y$ is usually close to, but not exactly equal to, $dy$ .

This notation is especially useful when several measurements contribute to an engineering calculation. It preserves units and makes first-order error propagation visible.

If $y=f(x)$ , define the input differential $dx$ as a chosen small change in $x$ . The corresponding differential in $y$ is

\boxed{dy=f'(x)\,dx}.

The actual output change is

\Delta y=f(x+\Delta x)-f(x).

For small $\Delta x$ ,

\Delta y\approx dy.

Guided walkthrough

Estimate a change in area

A circular plate has radius $10$ cm. Estimate the change in area if the radius increases by $0.03$ cm.

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Concept

Derivative as a local conversion factor

The relation $dy=f'(x)dx$ says that near a chosen input, the derivative converts a small input change into an approximate output change.

Modeling lab

Volume added by a thin coating

A metal sphere of radius $r=10$ cm receives a coating of thickness $dr=0.02$ cm. Since

V=\frac43\pi r^3, \qquad dV=4\pi r^2\,dr,

we estimate

dV=4\pi(10)^2(0.02)=8\pi\text{ cm}^3.

This is the surface area times thickness, exactly the geometric shell interpretation expected for a very thin coating.

After the explanation

Use the section idea

Reading lens

A differentiable curve behaves like its tangent line over a small enough neighborhood, with concavity explaining the direction of error.

Mental model

Linearization, differentials, and Newton's method reuse one local line for estimation, uncertainty, or a better root guess.

Decision

Choose a nearby easy input, record the local slope, and state why the requested change is small enough for the model.

Common trap

Presenting a tangent estimate as exact or running Newton iterations without checking residuals and failure modes.

Check yourself

Can you give the estimate, error direction, units, and a reason the local line is trustworthy here?

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