**9.9 Small Changes and Approximations**

$$\text{If}\delta x\text{isverysmall,}\frac{\delta y}{\delta x}\text{willbeagoodapproximationof}\frac{dy}{dx},$$ ,

This is very useful information in determining an approximation of the change in one variable given the small change in the second variable.

**Example:**

Given that

*y*= 3*x*^{2}+ 2*x*– 4. Use differentiation to find the small change in*y*when*x*increases from 2 to 2.02.

*Solution:*The small change in

$\begin{array}{l}\frac{\delta y}{\delta x}\approx \frac{dy}{dx}\\ \delta y=\frac{dy}{dx}\times \delta x\\ \delta y=\left(6x+2\right)\times \left(2.02-2\right)\to \delta x=\text{new}x-\text{original}x\\ \delta y=\left[6\left(2\right)+2\right]\times 0.02\\ \text{Substitute}x\text{withtheoriginalvalueof}x,\text{i}\text{.e}\text{.}2.\\ \delta y=0.28\end{array}$
*y*is denoted by**δ**while the small change in the second quantity that can be seen in the question is the*y**x*and is denoted by**δ**.*x*