When are differentials actually useful?

Question

I think of differentials as a way to approximate $\Delta y$ in a function $y = f(x)$ for a certain $\Delta x$.

The way I understood it, the derivative itself is not a ratio because you can't get $\frac{dy}{dx}$ by taking the ratio of the limits of the numerator and denominator separately.

However, once you do have $\frac{dy}{dx}$, you can then think of $dx$ as $\Delta x$ and of $dy$ as the change in $y$ for a slope $\frac{dy}{dx}$ over a certain $\Delta x$.

My problem is that I don't know why differentials are useful. The examples I saw are along the lines of approximating the max error in a sphere's volume if we know that the radius we're given (let's say 20) has a possible error of let's say 0.01.

In this kind of example, it seems to me we're better off computing $V(20.01) - V(20)$, instead of $\Delta V \approx dV = V' \cdot \Delta x$.

In the first case at least we get an exact maximum error instead of an approximation.

So my question is: When are differentials actually useful? Are they ever anything more than at best a time saver over hand computations? Are they useful at all in today's world with Wolfram Alpha and other strong computational engines around?

Answer 1

The list is endless, as the comments have started to indicate.

An example of an extremely important real-world application, on which literally trillions of dollars a day depends, comes from the risk management (hedging) of derivatives contracts from the point of view of a large bank or dealer. Say you call up Goldman Stanley's equity desk to purchase an equity call option, where the underlying equity has a continuously quoted market price of $S_{t}$ and the pricing function for the derivative contract is $f(S_{t},t)$ (at least according to the pricing models the dealer depends on and uses), then the "delta" of the derivative is $$\frac{\partial f}{\partial S}(S(t),t)$$ and represents a first order approximation of the amount of risk a dealer has from selling the option to their client (i.e., as a first approximation, it quantifies how much money the dealer will gain or lose when $S$ changes by an amount $\Delta S$). In particular, $$\Delta f_{t}\approx\frac{\partial f}{\partial S}\Delta S_{t}.$$ The market-maker is thus constantly buying and selling $|\partial f/\partial S|$ of the underlying asset in order to hedge (eliminate $S$-risk of) their exposure. They make money from the initial transaction costs/spreads, provided they effectively execute the hedging strategy, and avoid taking bets that their short position will be in the money and their clients' long position out of the money.

Obviously this differential is only a first-order approximation (and we're ignoring dependence on $t$ as well). To neutralize risk even more effectively, traders also try to be "gamma-neutral" in addition to "delta-neutral" by adjusting their hedging strategy according to the second-order approximation $$\Delta f_{t}\approx\frac{\partial f}{\partial S}\Delta S_{t}+\frac{1}{2}\frac{\partial^{2}f}{\partial S^{2}}(\Delta S)^{2}.$$

Other differentials that traders commonly use to manage their risk involve the quantities "vega", "charm", etc., these nicknames just being cute trader-speak for the partial derivatives of the pricing function with respect to other market variables for which it depends, including volatility, time, etc., respectively.

Answer 2

A pre-calculus answer What if you are asked of a ratio? For example, the marginal income tax rate. Let $T(x)$ be the tax you need to pay, where x is your income. You can of course, print out all the possible values on stack of papers - $$T(1) = 0.1, \space\space T(2)=0.2,\space\space ...\space\space T(10000) = 1100,\space\space T(10001) = 1100.15,\space\space ...\space\space T(50000) = 4600,\space\space T(50001) = 4600.25,\space\space ...$$ What a mess! It's much easier to say $$T'(x) = 10\% (x < 8000), 15\% (8000 < x < 30000), 25\% (x > 300000)$$. Plus, sometimes, the ratio is all people care about. For example, when the police is giving you a speeding ticket, it doesn't matter how many miles you drove, but it is the speed (differential between distance and time).

A calculus answer In some cases, in order to get an accurate answers, you need to do integrations on the ever changing differentials and solve equations when the relationship is inter-dependent. For example, when you are trying to how much distance will a space rock cover before it burns out (therefore determine whether it will hit earth), the velocity affects the heat generated, which affects the size of the rock, which affects the surface area of the rock, which in turn affects the velocity. This problem is insolvable if the concept of $\frac{dy}{dx}$ were never introduced.

Answer 3

Optimization, function analysis, computations (computers need a general approach unlike humans), financial and economical applications, engineering, simplifying expression.... the list is literally endless.