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:

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 $dr=0.02$: