Why does the fractional/relative change concept by differentiation $dx/x$ not work on big changes?

Question

There's a term that I've studied about called 'Fractional/relative change'. Basically, if say:
$$P = x^ay^bz^c$$
then, $$\log P = a\log x + b\log y + c\log z$$ By differentiation w.r.t $dp$:
$$\frac 1 P = \frac a x \frac {dx}{dp} + \frac b y \frac {dy}{dp} + \frac c z \frac{dz}{dp}$$,
$$\frac {dP} P = a \frac{dx}{x} + b \frac{dy}{y} + c \frac{dz}{z}. $$

If we increase any of the terms from x, y, z by any magnitude of percentage, then this formula should give the percentage change in 'P'. But, that only works for small changes, not for large ones, Why?
For example: $KE = (1/2)mv^2$. If the velocity is increased by 2%, what will be the fractional change in K.E?
To find this, shortest method would be to do this (in my opinion):
$$\frac{d(KE)}{KE} = \frac{dm}{m} + 2 \frac{dv}{v}$$ (acc. to fractional change formula).
So, $\frac{d(KE)}{KE} = 0 + 2\times2 = 4 \% $, which is correct.
But, if the same velocity is increased by 10%, then the change in kinetic energy goes up to more than 20%. As the change in velocity gets bigger, the change in K.E gets more than the double of the percentage change. I don't exactly get why does this happen. Even if it's a bigger increase and 'dx' is only used for small changes, it's eventually made up of small fractional changes that adds up to get a big increase. Why doesn't it follow the same pattern at bigger changes?

Answer 1

The relative change value is based on the derivative and assumes the higher order terms are negligible. You can see it with the simpler $y=x^2$. We have $\frac {dy}y=2\frac{dx}x$. This is valid as long as $dx$ is small and neglects the term in $(dx)^2$. As the change in $x$ gets larger, the term in $(dx)^2$ is no longer negligible. The local approximation follows the tangent to the curve, which is not such a good approximation a ways away.

Answer 2

One of the reasons for calculus in the first place is that, if $dx$ is infinitesimal, it makes the results a lot easier. Let's take $y = x^{100}$. The derivative is simple: $\frac{dy}{dx} = 100x^{99}$. However, if $dx$ is not infinitesimal, the result actually has about 100 terms!

For a non-infinitesimal $dx$, you would solve it like this:

$$\frac{dy}{dx} = \frac{(x + dx)^{100} - x^{100}}{dx}$$

You would need to use the binomial theorem to solve this. However, it would give you a correct result for non-infinitesimal $dx$, but would have a huge number of terms.

There is a study of calculus using finite $dx$s, sometimes called "discrete calculus" or "difference equations" (as opposed to differential equations). Many (most?) books on discrete mathematics has at least some treatment of this.

The reason why the large changes don't follow the pattern of the small changes is because the slope is continually changing through the region. So, on a small change, the slope hasn't changed significantly enough to cause problems. However, on a large change, the slope itself is radically different by the end.

Mathematically, remember that these extra terms all have some power of $dx$ attached. This means that, for really small changes, these extra terms are effectively zero. But, for higher changes, they actually influence the result.