Answer
$\Delta y \approx \frac{dy}{dx} \cdot \Delta x $
Work Step by Step
Step 1 In this problem, we have to explain the use of differentials in the approximation of errors and increments. The main idea is that if you know the relationship between $y$ and $x$ such that $\frac{dy}{dx}$ at a point is known, then for small deviations of $x$ from that point, the corresponding deviations in $y$ can be found.
Step 2 Consider the function $y = f(x)$, and increment/error $\Delta x$ from $x$ to $x + \Delta x$. Then we can write \[ \frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} \quad \text{(Equation 1)} \] The above equation is true for small values of $\Delta x$ because of the following definition: \[ \lim_{{\Delta x \to 0}} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} \] We can use Equation 1 to find the corresponding change in $y$ as follows: \[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \] In order to use the above equation, the relationship between $y$ and $x$ must be known. Result \[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]
