A Playful Trial of Derivatives
Author: Yi An (Another essay for a contest)
Derivatives are a truly wonderful class of mathematical tools. Her beauty lies not in the twisted and inconceivable calculations of tedious conic sections, but in the mutual combination of a few simple, plain formulas. Just as the Big Bang generated the entire vast universe from a single singularity, she gives birth to detailed spaces that cannot even fit within the entire answer sheet.
However, before commencing the statement, I wish to first present another method of proof regarding Vieta’s formulas, which is similar to the method used when solving many derivative problems.
This has exactly the same effect as using Vieta’s formulas when the difference between the highest and lowest degrees is exactly 2, while in specific operations, it saves the time of converting to a common denominator. Next is the proof regarding the product of the two roots in Vieta’s formulas, corresponding to the above, using the form of adding two equations.
Next is a derivative problem that I privately consider extremely typical, involving the addition and subtraction of related expressions of zero points, as well as the form, coming from the 2021 Guangzhou City General High School Graduating Class Comprehensive Test (1).
First, obtain two expressions
What is the reason? The reason lies in not immediately noticing that x is one of the common factors. Even if complete factorization is impossible, when solving problems, we should ensure that we attempt to factorize simple expressions at all times. Meanwhile, due to the existence of lnx making x>1 hold true constantly, the equation remains valid after dividing both sides by x, thus yielding the following
At this moment, it seems we have fallen into a plight of helplessness again. But when we examine the problem again, we find that this inharmonious a does not appear a second time in the question stem. This perhaps indicates that we should utilize a to act as the bridge for our solution, for it truly has no other use. We attempt to construct an equation regarding a:
Presumably, after enduring the tribulations of the sea of questions, everyone present has long possessed a sensitivity deep within the soul regarding the expression on the right side—that is, to achieve the goal by constructing t as the quotient of x1 and x2 and using substitution. However, this requires that the degrees of the numerator and denominator be the same; for example, in the expression below, both upper and lower degrees are 2
However, since the degree of x in lnx1x2 is 0, if we use the substitution method mentioned above, it is bound to be incomplete.
At this juncture, we need to liberate our minds, seek truth from facts, and advance with the times; this is the living soul of Marxism. That is, to conduct a bold Aufhebung of lnx1x2. Since it is like the a in the previous text and is of no help to solving the problem, why not attempt to abandon it:
At this point, it is still difficult to establish the correctness of the step of discarding lnx1x2. However, with sensitivity as strong as yours, did the statement in the question stem immediately make you think of that legendary inequality chain? Here, I write out her two-dimensional form from memory
Harmonic, Geometric, Arithmetic, Quadratic: The basic inequalities are the middle two terms, while the content of the question stem is extremely similar to the fourth term, thus we can deduce
And how could such sensitivity be consolidated by merely a few hundred questions? Let us return to the text above, using t to represent the quotient
Under the extreme tension of the examination hall, presumably everyone might recite cos as sin in the law of cosines or mistake the abscissa for the coordinate, let alone this simple common denominator in the final step. My experience is to simplify term by term before combining, rather than drifting up in an instant. I have had a similar realization regarding such small tricks in conic sections, so I insert a piece of content related to conic sections here
However, the most difficult part of conic sections is fundamentally not the simultaneous equations. Quite the opposite, the simplest of the simple is combining the line and ellipse equations and finding the relationships between various roots. What comes after this is “the Road to Shu is harder than climbing to the blue sky.”
For the expression above, my suggestion is to calculate -2a2 first, then calculate km, and then multiply them, rather than foolishly calculating -2a2k after finishing -2a2 and only then -2a2km; the latter is prone to calculation errors.
Next, returning to the main topic, I only recently suddenly understood why it tells us the magnitude relationship involving multiples between the two roots. The fundamental reason lies in wanting us to divide! It indirectly hints at using the substitution method to replace their quotient. But as I am dull and obtuse, how could I have completed the problem entirely? It is easy to know
Usually, for cases like this where a zero point seems to exist but cannot be found, rest assured and boldly use e, e/2, 1, 2, 1/e to substitute into the variable. The question setter went through great hardships just to make it make up zero exactly at that number. Thus
Let us look back for the nth time,
The veil of night is heavy, the cold crows caw repeatedly; overcome with weariness, I hereby lay down my pen.
Presented on the night of May 14, 2022
Because I started writing in google docs, and the formulas were done by casually downloading a plugin, which then seemed to convert them into images, I just brought them over in image form for convenience. I just looked up mathematical formula representation in hexo and markdown; it seems KaTex is not bad. I feel like I can put some math stuff up in the future, and incidentally practice representing mathematical symbols with KaTex. I feel this will also help improve my own problem-solving ability. After all, what is discussed above is such a minuscule part of the content.