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You have saved me throughout my exams over the years, thank you bro! I’m a 4th year in college and it looks like I’m gonna be staying a while longer. Thank you for doing what you do man
7:22: This is unnecessarily complicated! The derivative of e^x is e^x. You learn this in school right at the beginning of differential calculus. You don't need to use the extended exponential rule specifically for this.
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Snouldn’t it be d/du in your aside derivative of a^u? It’s not just about “any variable”, it’s about having the variable we’re deriving over in the exponent. If we did d/dx[a^u], u would just be some constant in terms of x. So it’d be d/dx [a^u] = u * a^(u-1) just like if it were some normal number. Only if we have d/du [a^u], or d/dx [a^x], is your solution correct. Somebody correct me if I’m wrong.
I'm super duper late but in case somebody reads the comment I'll explain why. You say u in dx is just a constant, that's right, but then d/dx of a^u would just be 0 since a and u are both constant. It would only be u • a^(u-1) if a is x. Now on why he used a^u on the example but d/dx on the solution. For me, we use u for the rules of derivative to indicate that u' is also part of the rule like sin(u) is cos(u) • u' (or du) if we used x then x' or dx is just 1 and sometimes the value inside the function is 3x, x², etc. not just a simple x. So to avoid confusion and to include whatever x you have there (3x, x², etc.) they use u to stand as a function instead of x. If my explanation is not clear in some parts, feel free to ask for clarification. I don't really know how to properly explain it through comment but I tried my best
No! Quotient rule and de L'Hospital's rule are two different tools for two different use cases. One rule does NOT replace the other! The quotient rule is always used when a fractional-rational function, i.e. a fraction, is to be derived as a whole. In the cases in which de L'Hospital's rule is applied, you are also dealing with a fraction, but you do NOT want to derive it, but rather determine its - if any - limit value. If both function terms in the numerator and denominator result in zero (at the same time), then and only then can you change the given fraction in such a way that the numerator and denominator are derived separately from each other.
To reassure you, I can assure you that the division 0/0 (or infinity/infinity or the other five indefinite forms) is and remains indefinite! (That's why they are usually put under quotation marks: "0/0") This fact does not change by applying de L'Hospital's rule. You may think that this rule legalizes something that is forbidden. She doesn't do that. But when it is used in examples, the language is often sloppy, which obscures the exact circumstances. The following happens: In most of these examples we are dealing with a fraction of two terms: y(x)=p(x)/q(x). Both terms p and q are functions of the same variable x, and unfortunately both terms become zero at the same time, for example when x takes on the specific value a. At x=a the fractionally rational function y=p/q has a gap in definition. It is not defined there. If de L'Hospital's rule (or some other trick) still gives a value y=b for the fraction p/q at x=a, then that doesn't mean that the indefinite form "0/0" now has a defined value had received. It still remains undetermined. But the concrete function y can be “repaired” at the point x=a. It is also said that the point of discontinuity can be lifted (or remedied) there. At this point you can now supplement or continue the given function with a concrete value b. You then write: y(x) = ... to the right of that you put a curly bracket and write two lines. In the first line you write the given function p(x)/q(x), and that it only applies to xa. In the second line you write the value b and that it replaces the function term in the first line for the case x=a.
Let's generalize the matter: y(x)=(a^x - b^x)/x. Let a>b>1. If x->infinity, then we are dealing with the form "(infinity minus infinity)/infinity". In this situation, the L'H (de L'Hospital's rule) is initially not applicable at all, because the numerator itself is already an indefinite form "infinity minus infinity". If we wanted to determine the numerator alone, we can use L'H. But first we have to change this indefinite form to the form "0/0" or "infinite/infinite". Then we can apply L'H. But unfortunately that doesn't help, because applying this rule leads to a modified term that is even a little more complicated than the original term. But through other considerations you can find out that the numerator becomes infinite. With this knowledge we could finally apply the L'H to the entire fraction term. But unfortunately that doesn't help, because the result is the same as before. Instead, the limit of the fractional term can be found in another way: We have assumed a>b. Therefore a^x > b^x for all x. We therefore reduce the entire fraction by a^x: lim (a^x - b^x)/x = lim (1-((b/a)^x)/(x/(a^x)) = ... For x->infinity the two terms (b/a)^x and x/(a^x) approach zero: ... = lim (1-0)/0 = infinity. .//
3:35ff: Really? 9-digit decimal numbers? And then, especially, 5:00ff: It is completely nonsensical to specify five-digit (!) decimal numbers (in the table) if the crucial changes that you want to demonstrate only occur in the first two digits after the decimal point. I quote Carl Friedrich Gauss: "Nothing shows the lack of mathematical education more than in an exaggeratedly precise calculation." Modern media education could, among other things, also consist of not blindly copying all the numbers that the pocket calculator gives you.
