Differentials are about how very small changes in input can affect the output
$\Delta x$ is the change in $x$, $dx$ is an infinitesimally small change in $\Delta x$
$\Delta y$ is the change in $y$, $dy$ is an infinitesimally small change in $\Delta y$
What you might notice is the $dy$ and $dx$ that is seen in Leibniz Notation
And also divding both changes, $dy$ and $dx$, looks like the slope formula, which is a derivative
$$\begin{align} \frac{dy}{dx} = f'(x) \rightarrow dy = f'(x)dx \end{align}$$
This also means that $\Delta y$ (which is normally $\Delta y = f(x + \Delta x) - f(x)$) can be found easier using differentials
If $\Delta x$ is small, then $\Delta x \approx dx$ and $\Delta y \approx dy$
Practice Problems
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Find $dy$ and $\Delta y$ if $f(x) = e^{-x} + cos(x)$ as $x$ goes from $1$ to $1.02$
$$\begin{align} f(x) &= e^{-x} + \cos(x) \\ f'(x) &= -e^{-x} - \sin(x) \\ \Delta x &= 1.02 - 1 = .02 \\ \Delta y &= f(1 + .02) - f(1) \approx .883961 - 0.908181 \approx -0.02422 \\ dy &= -e^{-1} - sin(1) \cdot .02 \approx -.024187 \end{align}$$
You can see that the values are pretty close
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The radius of a circle is $9$ in with a margin of error of $.03$ in. What is the max error in the area of the circle?
Using area of circle formula
$$\begin{align} f(r) &= \pi r^2 \\ f'(r) &= 2\pi r \end{align}$$
The margin of error is our $\Delta r$, and since it is small it can be approximated to $dr$
$$\begin{align} \Delta r &= .03 \\ dr &= 2\pi (9) \cdot .03 \approx 1.69646 \end{align}$$
While this seems like a large error, if we take it in respect to the area of a $9$ in circle it's not so big
$$\begin{align} f(9) &= \pi 9^2 \approx 254.469 \\ \frac{1.69646}{254.469} \approx .00667 \rightarrow .0000667% \end{align}$$
So in reality it's a small error