~Here is a brief essay on the differential & how it can make rate of change problems much easier to compute.

~The forms of the chain rule depend upon treating the derivatives involved as fractions with "free" numerators & denominators. For example, dy/dx can be treated as a separate quantity, dy, divided by a separate quantity, dx. So, when we multiply & divide derivatives involving them, we can treat them as separate entities. i.e., (dy/dx)(dx/dt)(dt/ds) = dy/ds or when dividing, dy/du divided by dx/du we get dy/dx.

~We have not justified treating those quantities separately. That's the purpose of this essay. The key is to define the top (dy) and the bottom (dx) as separate quantities.

~This is not in the course outline & should be covered for justification of the uses of the chain rule & rate of change problems.

~As you recall, dy/dx represents the slope of the tangent line to a curve at a given point. To motivate the definitions, we want to define these quantities separately in such a way as to be consistent with this slope.

~So, here's what we do: define dx "the differential of x" as any real number. So, this means that dx could be any change in x. It will be considered an independent variable. (i.e., any delta x or h used in previous limits) no matter how small or large.

~Once we define dx, we can then define dy (a dependent variable) as a function of the derivative,  
f ' (x), and dx the following way: define dy "the differential of y" as a function of f '(x) and dx by:
dy = f ' (x) dx.

~Notice by defining dy this way, the ratio of dy to dx is consistent with the derivative & still gives the same result as before. i.e., dy/dx = [f ' (x) dx] / dx = f ' (x).

~Now, we have the right to separate the top & bottom of the first derivative, thus the chain rule can be applied & other rate of change problems can be done.

~First, let's compute some differentials.

~Find the differential of y for y = 3x²-5x+7.  dy = f ' (x) dx by definition, so, dy = (6x-5)dx, done.

~Compute d[(e^x)(sin(x))].  Ans: {[(e^x)(cos(x))+(sin(x))(e^x)]} dx. Note: the quantity in front of the dx is always the derivative.

~Find ds for s = [sin(t^(-3))]².   ds={2[sin(t^(-3))][cos(t^(-3))][(-3)t^(-4))} dt.

~Also, rate of change problems are much more simplified when differentials are used.  For example: Find the rate of change of x³ with respect to x² at x=2. Using differentials, we can write d(x³) / d(x²). Then computing top & bottom, we get,  [3x²]dx / [2x]dx or 3x/2 which is 3 at x=2.

~This differential technique greatly simplifies the notation.

~So, in summary, if you would like the rate of change of quantity A with respect to quantity B,just compute dA/dB . (i.e., compute the differential of A then divide it by the differential of B).

~Another important use for differentials is to approximate the change in y of a function, i.e., using dy (change in y along the tangent line) to estimate Dy (the actual change in y between 2 points on the curve), for a small change in x (dx). This is important in estimation & approximation.

~Hope you understand this important essay.