Approximations Using Differentials

The idea

Near a point where $f(x)$ is already known, the differential $dy = f'(x)\,dx$ approximates how much $f$ changes for a small change $dx$, without needing the exact value:

Why it works: dy and Δy

For a step $dx$ along the $x$-axis, the curve itself rises by $\Delta y = f(x + dx) - f(x)$, the exact change. The tangent line at $x$ rises by $dy = f'(x)\,dx$, the estimate. The approximation replaces the curve by its tangent for that short step.

The gap between $\Delta y$ and $dy$ shrinks much faster than $dx$ does, which is why the estimate is good for a small $dx$ and poor for a large one.

Worked example: approximating a square root

Use differentials to approximate $\sqrt{36.6}$.

Solution: Take $f(x)=\sqrt{x}$, with $x=36$ (a perfect square close by) and $dx=0.6$. Then $f(36)=6$ and $f'(x) = \dfrac{1}{2\sqrt{x}}$, so $f'(36) = \dfrac{1}{12}$:

Worked example: percentage change in volume

Find the approximate change in the volume of a cube of side $x$ meters caused by increasing the side by 3%.

Solution: $V=x^3$, so $dV = 3x^2\,dx$. A 3% increase means $dx = 0.03x$:

Worked example: error in a measured quantity

A sphere's radius is measured as 7 m with a possible error of 0.02 m. Find the approximate error this causes in the computed volume.

Solution: $V = \dfrac{4}{3}\pi r^3$, so $dV = 4\pi r^2\,dr$. With $r=7$ and $dr=0.02$:

Common mistakes

• Choosing a far-away base point. Pick $x$ where $f(x)$ is easy and close to the target: 36 for $\sqrt{36.6}$, not 25.
• Using degrees in a trig approximation. $dx$ must be in radians. For $\sin 31^\circ$ from $\sin 30^\circ$, $dx = 1^\circ = \dfrac{\pi}{180}$, not 1.
• Mixing up error and relative error. $dV$ is the error in the volume. The relative error is $\dfrac{dV}{V}$, and the percentage error is $\dfrac{dV}{V} \times 100$.