Calculus Teacher vs. Power Rule Student
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- Опубліковано 8 жов 2021
- Definition of derivative vs power rule! Want to learn more about calculus with visual and physical intuitions? Head to Brilliant brilliant.org/blackpenredpen/ (20% off with this link)
The definition of derivative is the foundation of differential calculus. However, many students tend to ignore it and want to use the "shortcuts" (differentiation rules such as the power rule). Not only do we need the definition of derivative whenever we need to determine the derivative of a function, but I also give a calculus 1 problem where using the definition of derivative is easier than using the differentiation rules!
Check out "when d/dx(x^2) becomes the hardest question on your calculus final": 👉 • calc 1 final be like (...
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Check out "when d/dx(x^2) becomes the hardest question on your calculus final": 👉 ua-cam.com/video/wUD7n_ohMQk/v-deo.html
my brain stop working i dont get it
Ur fan sir
“The pen is blue”
~ Jim carry
I felt so smart in calc when I figured out the power rule on my own while doing the homework in my class but then the next day my teacher had a lesson on it and everyone else learned it too.
That’s great you figured it out on your own. I remember some guy in my AP calc class yelled out the power rule before I could even try to figure it out.
Gatekeeping formulae? Jk😂
SAME
It hurts man. I can feel. 💔
Yeah similar kind of thing happened in my class too
In 9th class
i knew structure of dna and other stuff related to that but then in 1 class our teacher taught everyone that, it hurted man
Because before that only i knew and i felt smart before that(btw structure of dna is basically 11th class topic...)
Math teacher: spends 1 hour explaining a method
Math teacher: and,there is an easy method too
😂 😂 😂
Lmaoo it do be like that doe
💗
Easy method's only good depending on the situation and function. Useless in Proofs, but convenient for applied calculations.
And then there's the feeling of proudness once you master the hard method
5:55 I laughed at that facial expression, "can we use L'hopital's rule?"
Impressive! I know this too, you know
😂
@@blackpenredpen nice beard😉
@@blackpenredpen I am here to serve beard cut for those who do not have beard cut themselves.
I don’t get the joke can someone explain?
@@colleen9493 watch from 2:10 for like 20 seconds.
You can solve the (2^x-1)...(2^x-10) with the product rule too, namely (fgh)' = (f')gh + f(g')h + fg(h'), etc. In this case all terms except the first one vanish and the first term is -(9!)log(2).
The students always win!!😃
Can you prove this with the limit defination though? That is the whole point of the limit definition is to prove these "shortcuts".
@@liamwelsh5565 I'd prove the multiple product rule by induction from the usual product rule. I guess you could prove it directly with the limit definition but instead of adding and subtracting just f(x+h)g(x) you'd need an extra pair for each extra function if that makes sense
Would be shorter in a sense to just make the first term f and the other 9 terms together to be g I suppose
If you did this, you would get zero points, as only your f is differentiable at x=0. The rest of them are NOT differentiable at x=0!
I.e. you cannot apply the product rule at all!
Funnily enough, when I first encountered the formal definition of a derivative in a real analysis textbook at university, my first question was why I was required to spend so much time in high school memorising the rules when just working from the definition was intuitively so much easier. Hindsight is often very ironic like that.
This right here is why high school calculus shouldn't count for college credit. Sure, you're technically doing calculus, but I could probably get my 12 year old nephew to do that much. Part of the point of the classes is that you've got at least some understanding of why the results are what they are, so if something goes wrong, you have some hope of knowing and fixing it.
@Autistic Doing and Thinking idk that seems like it only happened in your highschool calculus class.
In my ap calculus class we learned all the definitions and stuff, did the derivatives once and then just memorized our results so we didn't have to redo the work again.
Honestly Calculus is what made me want to be a mathmetician. I cannot explain why, but integrals were so seemingly magical to me as well as differentiation. Hope I'm smart enough to get my PhD in applied mathematics after my Bachelors!
Student wants to use the power rule? "Oh that's brilliant! How did you come up with that?"
If they don't know how it's derived - "that's why we're practicing using the definition of the derivative, so we can use the definition to prove interesting patterns we might notice."
If they do know how it's derived - "that's great that you've already learned some of this! I don't want to jump ahead so we can give other students time to think and process. But maybe consider yourself a resource for other students in this first unit!"
Absolutely, that’s a great method of probing your students’ advancedness without being judgmental toward either them or the other students.
I'm big-time guilty of this lol I read the text the night before the lecture so I'm usually the most vocal in my calc II class. I feel bad at times because I'm trying to push myself further in math and I'll ask questions that are a bit more complicated than the rest of the class is at, but at the same time I have people asking how to differentiate 3x (an actual question we had about 4 weeks ago, do not ask me how they made it to calc II) and I don't feel as bad
The fact that he just marked that timestamp as "final words"!!😆
Inaccurate, calculus teachers actually just throw a bunch of formulas at you, bprp is just built different.
It seems like you just had some had calculus teachers
both of my parents who were professors in china hate when I use derivative rules for a basic function
@@skylardeslypere9909 It appears most calculus teachers are bad calculus teachers. Tee hee.
But in all seriousness, the reason most calculus teachers are bad is because the recommended curriculum for calculus courses is itself garbage. Calculus is taught way too much like they are teaching how to do real analysis in every wrong way possible to get the right answer, rather than actually teaching how to it the right way.
@@user-lh5hl4sv8z You probably shouldn't be using the derivative rule for basic functions, just like bprp mentioned he would also just use the power rule if he was randomly asked on the street. But you should be familiar with the mechanism behind it, so you are not stuck when you see a different function that haven't learnt a formula for yet.
Interesting, neither of the Calculus teachers I had did that. Later on, they would, but we already knew the definition of the derivative and could generally do that on our own if we needed to. It got a bit more annoying at times with transcendental functions, but it could generally still be done with the use of the proper identities.
Whoa that is amazing!!!!!
Impressive, I know that too, you know?!
WHOAAA
Q: Is it possible to have a question so that it is actually easier to solve with the definition of derivative rather than the differentiation rules, which limit is actually easier to solve using the epsilon-delta definition rather than the limit rules?
The problem is that to use ε-δ you already have to know the limit. Which begs the question, how did you find that limit?
@@skylardeslypere9909 Maybe you have intuition about the limit (which allows use of epsilon-delta) but don’t know how to actually calculate it using the limit rules? 🤔
@@nicolastorres147 Or, I guess, you could use numerical methods to find the limit (well, guess the limit). I'd highly suspect that in the case you suggested, the limit would have to be zero or something simple
Linear approximation?
@@skylardeslypere9909 *The problem is that to use ε-δ you already have to know the limit.*
You absolutely do not have to know the limit to find it using the ε-δ method. For example, consider f : (-t, t) -> R with f(x) == x, and let x -> 0. There exists a real L such that for every real ε > 0, there exists a real δ > 0 such that for every real x in (-t, t), 0 < |x| < δ implies |x - L| < ε. |x| = |(x - L) + L| >= ||x - L| - |L||, which means that -δ < |x - L| - |L| < 0 or 0 < |x - L| - |L| < δ, implying |L| - δ < |x - L| < |L| or |L| < |x - L| < |L| + δ. |x - L| < |L| or |x - L| > |L|, meaning |x - L| is not |L|. This means either x is not 0 or x is not 2·L, or else L = 0. But the former is impossible since 0 < |x| < δ implying |x - L| < ε must hold true for x = 0 since 0 is an element of (-t, t). Therefore, L = 0, if L exists. Of course, this was tedious, but it was still possible to find L without knowing L a priori, using only properties of absolute value inequalities.
3:42 😂😂😂 he was just like “you get my point, there’s a shitload of rules, no need to list more”
It's 2:11 AM. I haven't taken a math class in roughly 15 years. I've never taken a calculus class. I have no idea what you're talking about. I have no idea why UA-cam brought me here. With all this being said for whatever reason...I LOVE IT. I guess the tube knows me better than I know myself.
To grow in mathematical maturity, it is necessary to look well beyond the equations. They fall naturally into place from the structure we are exploring: the mathematical scaffolding of the equations themselves. When I was 16, I was in such a hurry. I could certainly have been described as THIS student, and I just wanted the formula. Today, I struggle to make certain formulas stick in my memory, aside from the ones I learned much earlier in my youth. It leads to occasional moments of embarrassment in front of scholars smarter than I (by that I mean all of them); I mainly shrug it off, figuring I can derive them again from the landscape I've already explored, or look it up in a book. It's no problem for me. The equations come naturally from a deeper understanding of the structure I've studied. And more often than I realize, when I wander back through well-studied material, I find new perspectives to look at old things. What is old truly is new.
You perfectly encapsulated the art of mathematical study: we study the structure of truth, not the formulas which state a truth. The argumentative structure is what gives life to the mathematics. This fact is too deep for 99.9% of math students to realize until they get about half-way through a physics or maths degree, unfortunately. If we could teach children in Algebra 1 this fundamental and beautiful fact of mathematics, we may very well have millions more mathematicians and physicists than we do presently.
As a former math tutor, I find that with the "harder" derivatives, I break this out and it allows me to solve the problem with ease. It also helps me to better understand limits.
Nice video!
Try f(x) = 0 if x = 0 and f(x) = x²sin(1/x) for x not equal to 0
The derivative for this function isn't continuous and hence normal differentiating techniques don't work...we have to go by the definition of the derivative
@ridayesh parab
Another famous example where we HAVE to use the definition of the derivative is the following function:
f(x) = 0 if x = 0, else f(x) = e^(-1/x^2)
If you have a graphing calculator, take a look at it's graph near x = 0. Zoom in, zoom in...
You'll get the picture: the function is very flat near x = 0.
We will see what this means for the Taylor series approximation for the function f later on.
Let us first concentrate on the derivatives of f for x ≠ 0.
Since f is a composition of two differentiable functions, it is differentiable (for x ≠ 0).
We get:
f'(x) = (2/x^3) * f(x) if x ≠ 0
For the second derivative of f we find:
f''(x) = (4*x^-6 - 6*x^-4) * f(x) if x ≠ 0
Let's do it one more time:
f'''(x) = [ 8*x^-9 -36*x^-7 + 24*x^-5 ] * f(x) if x ≠ 0
(I hope I haven't made a mistake here)
By induction we prove that:
f(n)(x) = P(1/x) * f(x) if x ≠ 0 (*)
Here f(n) denotes the n-th derivative of f and P is a polynomial (the degree of P is irrelevant for now).
Note that the formula above states P(1/x), not P(x)!
(for the 3rd derivative, the polynomial P(x) equals 8*x^9 - 36*x^7 + 24*x^5, but remember we have to plug in 1/x, not x)
We MUST use the formal definition of the derivative to determine whether the first derivative of f exists for x = 0:
f'(0) = lim (x→0) (f(x)-f(0))/(x-0) = lim (x→0) f(x)/x = 0 (**)
since f(x) goes to zero much more quickly than x does if x→0.
Since lim (x→0) f'(x) = 0, we now know that f' is continuous everywhere.
Again, by induction, we can proof (using (*)), that all derivatives of f are continuous, differentiable and that f(n)(0) = 0 for all n.
(again, we HAVE to use the formal definition of the derivative here as we did in (**))
So we now know that f is infinitely many times differentiable for all x and that f(n)(0) = 0 for all n.
What does this mean for the Taylor series approximation of f?
Well, actually it means that every finite Taylor approximation for f must be the polynomial Tn(x) = 0!
Thus, here we have a (relatively simple) function for which it's Taylor series is NOT equal to the function itself: the function f is SO extremely flat near x=0, that only the zero Taylor polynomial/series can approximate it well.
Not every function equals it's Taylor series: the fact that the function is infinitely many times differentiable is not a sufficient condition, neither is the fact that the Taylor series is convergent for all x... It needs to converge to f(x).
It’s funny that when I was learning calculus in my spare time I didn’t know about the derivative rules so I would calculate the derivative by hand using the formula, I was so excited when I learned about the power rule and chain rule etc etc
Man, that limit of 2^h-1/h is pretty crazy man. Pretty triv w/ L'hopitals, but seeing that you used it in your search for derivs in your other video (and abusing the fact that you can get derivs of all the exponentials from just e^x) I tried to explicitly prove the case for e. Surprisingly tricky, and naturally one of those hard to search sorta math problems. Eventually managed it using epsilon delta and the logarithm definition of the exponential (ln(x+1) as integral from 0 to x of 1/t+1, actually the first rigorous def I was given in uni!) and the taylor expansion around 1+x of 1/x [which is gettable naively]). This was really helpful in taking care of the extra 1 in the limit numerator.
Pretty cool problem!
Your example can still be solved just as easily with the product rule.
Let g(x) = 2^x - 1, and let h(x) = (2^x - 2) ... (2^x - 9).
Then f(x) = g(x) h(x). So f'(0) = g'(0) h(0) + g(0) h'(0).
Since g(0) = 0, the quantity is just g'(0) h(0).
I am start to learning differential calculus in our school but your teaching and sums are really interesting thanks🙏
Hey, I've just discovered your channel and it's very cool. I've just finished what would be grade 11 in Spain (1r de Bachillerato) and I must tell you that I have learned to do the derivatives of functions using formulas for each case instead of using the general formula. For instance I have written down that the derivative of loga(x) = 1/(x*ln(a))
I am trying to become a professor at a university level as well. You videos are so motivating and relaxing at the same time
A perfect method, very interesting and the answer to the question of how derivatives are calculated. This is called also math culture. But ... in the middle of exam you use the table of derivatives ...
Excellent work.
It’s a good follow up question to ask how they know that and then use the binomial theorem to prove the general case where the fixed exponent is an integer.
Solution using general differentiation techniques (Modified Product Rule):
derivative of f(x) * g(x) * h(x) = f'(x) * g(x) * h(x) + f(x) * g'(x) * h(x) + f(x )*g(x) * h'(x) [This kind of pattern goes on for any number functions multiplied together]
2^0 = 1
So, all the terms that retain (2^x-1) will just become 0.
Hence, derivative of (2^x-1) * (2^x-2) * ... *(2^x-10) at 0 = (2^x-2) * (2^x-3)*...*(2^x-10) * [derivative of (2^x-1)].
I'm in highschool and my mathbooks and teacher just thought me the specific rules (like the power rule and multiplication and division rules and such) but never the general formula of making a derivative. I prefer knowing the formula over specific working outs of the formula even if I'll usually will still use the rules. So thank you
i really knew the fuction but i never used it
but now i can think of ways solving questions other than the normal rules
thanks man
I've actually been thinking about this question!
Logarithmic differentiation could also be used for the f(x) = (2^x-1)(2^x-2)(2^x-3)(2^x-4)(2^x-5). Rules of log would allow us to rewrite the product as a sum and knowing the derivative of 2^x, the rest becomes easy.
Familiarity with the simple way fosters and facilitates interest and understanding in the real way.
the question came to my mind but i dismissed it thinking that such a thing shouldnt exist because shortcuts will have to work always. but no !!
Amazing video and concept.
Oh..... and yeah my membership month is gonna run out soon . . . . gotta renew it asap !
My favorite UA-cam tutor.
Hi bprp, I've seen a recent video of Mathologer talking about calculus of differences, the ideas of calculus applied to sequences, where the integrals are sums and the derivatives are differences... That's so cool, I derived its "product rule" to difference and then the equivalent to integration by parts with the idea of get a closed formula for some sums like for example the aritmetic-geometric series and things like that... would be awesome if you can develop this ideas in a video! Much love from Spain
Thanks.
Have you seen the summation by parts? Prof Penn has a video on that ua-cam.com/video/mNIsJ0MgdmU/v-deo.html
@@blackpenredpen Wow I didnt see it... that's exactly what I was thinking, thank you very much! This is so cool and even maybe can help me with some series problems
In physics we also "discover" derivatives. E.g. If you are considering the pressure on a horizontal slice of fluid in a vertical tank, you end up with something like P(z+h) - P(z) = gravity*density*h. Then you know that if you divide both sides by h, the left hand side is just dP/Dz, so the pressure is the integral of gravity * density!
(h is just the height between the slice we are considering and a point a tiny distance above it.)
in the d/dx(x^2) case u can factor (x+h)^2 - x^2 as a difference of squares
Very nice! I point out, I think there might be an editing error at 3:43
“Impressive… I know this too, you know” 😭😭
I felt that
Linear approximation is even easier:
The linear approximation of 2^x about x=0 is 1+x*ln(2).
Let's set a = ln(2).
So we need to find the linear approximation of ax(ax-1)(ax-2)(ax-3)...(ax-9).
By the binomial theorem, or even by simple inspection, the linear term here is -9!a*x. The constant term is 0.
So we have a linear approximation of -9!a*x, the slope of which we can read off as -9!*ln(2).
two other uses : 1. in numerical analysis you might need to take a derivative using the definition. 2. If you are not sure about a certain derivative in an exam, you can use your calculator to check. In both cases you take h very small but not zero.
Thanks for teaching, now I learned on how to count from one to ten
h/h doesn't "Cancel Out," it reduces to one.
For instance in xy = yz, you can't "Cancel Out" the y because you can only "Cancel it" if y doesn't equal 0. So, saying "Cancel Out" is actually informal and saying reduces to one lets the reader/student know that the y doesn't just disappear yet that it goes to a value of 1.
My teacher was always super rigorous when it came to math.
That's nonstandard. h/h has the hs canceling, as the term is generally accepted to mean. You can cancel them because there's a removable discontinuity that we've already decided to ignore via the limit pushing the sides together. You can definitely cancel the y in that expression, you just have to be mindful about the possibility of y = 0. In that case, it doesn't much matter what x or z are equal to, as any value of either would satisfy the equality. So, you'd likely set a domain restriction on y and move on.
As a physicist, you almost never solve for a derivative using the definition (I at least have definitely never) BUT the definition is very important because there are several cases, classical mechanics and statistical mechanics especially come to mind, where you are solving a physical problem and end up with the definition of a derivative. If you never learned the definition you would not be able to solve it beyond that point. This is true in general rather often, where you rarely use the definition or derivation of a method to directly solve problems with that method but knowing the derivation lets you recognize it and transform your difficult problem into one that you already know the tricks to solve.
Hello sir, I am from India and I love the way that how easily you solve these questions in a funny way❣❣
what an amazing example!
You don't need the derivative definition to derive 2^x you transform it into the definition using 2^x=exp(X(ln(2)) and that's easy to differentiate.
I reckon that your 2nd example should apply linear approximation as you let f'(x) = f'(0). It seems like you take derivative of this function at x = 0, and you guys should recognize this approach should be more easier to handle since it will appear a bunch of zero.
Remarks:
It is just useful when x is really really close to zero but not x=0. According to def. of derivative, it just apply when one of the two points is close enough the other point.
We can do the above problem by direct differentiation
Simply take natural log of both sides then differentiate normally by chain rule and proceed to find the answer.
You might run into some problems with 2^x-1 but u can cancel it out with first term of the fractional part and all other terms proceed to become 0
Love you brother....God bless you !!!
As a physics student I have needed to use a variation of the definition of the derivative to derive certain equations in classical mechanics.
Ah yes
Open Matlab and load in the forward finite difference algorithm you had to write in like physics 105
Using differentiation we can make the same observation that any terms containing (2^0 - 1) will cancel and immediately arrive at the same answer with less effort.
Finding the derivative is much easier when you already know the derivative
When doing differentiation by first principles I don't like writing down the limit notation everytime either. This is why I was taught by my lecturer to obtain the Average Rate of Change(Δy/Δx) first and then once we get that we can obtain the Instantaneous Rate of Change(lim Δy/Δx as Δx---->0)
When I do problems for myself rather than formally I just use the " " for each successive limit after the first one. Ssssstttttt, Don't tell my old calc teachers
“The proof of this is left to the reader.” 😂😂
you could just do the question like this
f(x) = the given thing
we can just assume 2 new functions
g(x) = 2^x-1
h(x) = (2^x-2)(2^x-3).....(2^x-10)
so in this case f(x) becomes
f(x) = g(x)*h(x)
and here we can use product rule :D
f'(x) = g'(x)h(x) + g(x)h'(x)
and at x = 0
f'(x) = g'(0)h(0) + g(0)h'(0)
here actually, we see that g(0) is just 0, so it doesnt matter what h'(0) is cuz the final result would just be 0
we know that --> g'(x) = 2^x ln2 from differentiation rules, so g'(0) = ln2
and h(0) = (-1)(-2)(-3).....(-9) = - (9!)
so
f'(0) = (ln2)(-1)(9!) + (0)(h'(0))
f'(0) = - 9! ln2
it isnt gonna be faster than definition but with enough practice its faster, in my case i didi t like this way faster in my head
I’m in accelerated precalc right now and I can’t wait to get into this stuff
Start right now. Go through khan academy’s calc sequence. You don’t really need precalc before calculus if you have sufficient algebra skills.
That’s how I felt lol, I literally could not wait! Now I’m in my second year of calculus.
I studied calculus when I only knew algebra. Things like derivatives and integrals don’t really require all that until you get into partial fraction decomposition and trig sub. Good luck!!!😊
@@GPLB Actually doing that RN. The AP Calc and Limits are all understandable with Gr 10 precalc
My calc teacher showed us how all the different types of functions with the limit law and then allowed us to use the "shortcut formulas". However, we were asked to prove some of them on the exam and were still asked to use the limit definition for some of the questions.
@Liam Welsh
I'm curious how your calc teacher showed you that (e^x)' = e^x (let alone how he got to the definition of 'e').
You could use it's power series for that (which is already 'sophisticated'), but the fact that you can interchange d/dx with ∑ in this case should be proven...
Using only exponents, you can easily derive that for f(x) = a^x, the derivative (if it exists) must satisfy f'(x) = f'(0) * f(x), but how does he prove that f'(0) exists? And what is it's value?
What if we take log(.) On both sides and then differentiate it.
F'/F = ln2 (2^x) /2^x-1 + ..........
Now
F' = ln2 (2^x) {(2^x-2)(2^x-3) ..... (2^x-10) + ...........
Now
At x=0
Expect first term other terms will be 0.
And first term will be
-ln2 (9!)
I love it when he says something wrong and then kinda starts the whole video over because of that one blunder, instead of repeating the word.
I'm a highschooler currently learning calculus, this is amazing. Let me realize that the ahortcuts were not always available. Such an interesting concept.
If you've got time, find an online copy of a very old Calculus book from a hundred years or more back. There's a ton of things that are now standard that at the time were rather obscure or completely unproven that we can now use without having to prove them out every time.
Please make videos on sets, relation and functions
it is called Mathematics
Is there a problem where it would be easier to use the definition of multiplication (a*b = Σ(n=1,b) a) rather than multiplication rules? That's what this question sounds like. Knowing definitions might not be as practically important as learning how to use them but it's still an important part of understanding
I like how you put -(9!) instead of -9! for clarity.
Take this *9! ln1/2*
By the product rule you can find it easily too. The derivative of 2^x-1 is ln2 2^x which is ln2 at x=0 and times all the rest which is -9! At x=0 and then + 2^x-1 which is 0 at x=0 times the derivative of all the rest which we do not care case it is 0 times somthing which is 0
d/dx of 2^x is (ln2)e^(xln2), which shares only one point in common with x(2^(x-1)) when both are plotted. So using the power rule for 2^x doesn't work.
BPRP on the final part do you not have an even number of factors so should it not be (-1)^(even integer)*ln2*9! which just equals ln2*9!. I think everything else checks out but I was a bit hesitant or uncomfortable at first with commuting the negative signs around - somehow had a thought of don't you need to make a special definition of factorial for negative numbers just like for rational numbers for the gamma function etc. (probably thinking too hard for this part).
Please make a tutorial on how to switch markers like you :3
What about taking this a step further: are there any problems where it's easier to evaluate that limit using the definition of a limit rather than any of the shortcut rules?
Just great video!
in my country a^x where a>0 the derivative has a rule, (a^x) ' = a^x • lna that's why e^x's derivative is the same cause the lne is equal to 1
question: Isn't f'(0) the derivative evaluated at zero (so apply def of derivative first, then plug in 0)? What allows us to plug in zero during the process of applying the defof the derivative? some continuity reason?
f'(x) = lim [f(x + h) - f(x)]/h (h -> 0) is true for every x for which the latter exists. This necessarily implies f'(0) = lim [f(0 + h) - f(0)]/h (h -> 0). Your comment seems to suggest that "plugging in" is an operation in itself, but this is not the case.
Is it possible to make a definition of an integral by the definition of a derivative?
in england, in the exams if they ask you do find the derivative from first principles, you have to do it the long way to get all the marks
Beautiful!
I wonder if it is possible to introduce calculus via the Jacobian before the 1D definition. 🤔
Somehow I can feel his anger through the screen. As a student, I always loved the limit definition (It was reasonable and easy to understand). I remember when I tried to ask my teacher if I could use syntheic division instead of polynomial division, and she got reasonably angry with me because it was just a shortcut without knowing how it truly worked.
Is there a difference between lim h -> 0 (f(x+h)-f(x))/h ,lim h -> 0 (f(x+h/2)-f(x-h/2))/h ,lim h -> 0 (f(x)-f(x-h))/h ??
"Impresive, I know this too :) 2:26" 🤣🤣🤣🤣🤣🤣🤣🤣
😂
the face made me laugh so hard. I had that experience with my teacher. I ask her so many questions, abstractizating from the subject and she just gives me that face 😅🤣
We weren't taught derivatives at school, so when I started math studies, no one would say "use the power rule', we had to derive all the rules for elementary functions. When we had that calculated, we were allowed to use it.
My calc teacher really spent days going over this only to hit us with a hammer and tell us we bring the exponent down. Now here I am going back to square 1 but with polar coordinates in calc 2
Amazing!
dude you look like jordan rudess I love it
This is very useful because I have absolutely no idea what derivatives are!
I’m only learning limits now
I got a similar question when talking about Riemann sums, do you know of any integrable function where it's easier to use Riemann sums rather than trying to find an antiderivative?
Transcendental functions often times wind up being easier like that. The algebra involved often makes trying to use more conventional integration techniques untenable. As time goes by, there are fewer and fewer things that need to be done in that manner as more and more formulas have been proven to take care of certain insances.
In Newton's time, most integration was the result of integrating the series approximation for the function and over time better methods of integration were proven and adopted to reduce those situations.
There are plenty of functions that are integrable but whose integral is discontinuous and therefore not differentiable
Hello, we know that f(x) = 0 and f(x) = e^x follow the rule f(x) = f'(x) but any other f(x) follows that rule?
You should re-name the video to: That one student
Thank u from algeria 💕🇩🇿
I'm stealing that example
Be my guest! 😃
What is the difference of writing (f(X+h) - f(X))/h and (f(X) - f(Xo))/X-Xo ?
3:44 He Just gives up
Well actually here in France the teachers do it the easy way :/ And we got tons of different formulas to learn for different derivatives.
I burst out laughing at the 'yeah, impressive, I know this too you know' after the student explains the faster method lmao
That's insanely cool.
Hey Professor! Can you please tell me how can we differentiate for irrational exponents? As the binomial expansion is defined for rational exponents. So is there a sort of 'power rule' for that?
The power rule works for *all real exponents*.
The reverse power rule for integration, works for all real exponents, other than -1.
To prove the power rule for all real exponents, you can re-write the power function in terms of a base e exponential. We'll accept the derivatives of ln(x), exp(x), and the chain rule, as rules that have already been proven.
Given y = x^r, where x and r are both real numbers, and r is a constant, while x is a variable. We are interested in finding dy/dx.
We can re-write in terms of base e:
y = exp(ln(x) * r)
Take the log of both sides:
ln(y) = ln(x) * r
Take the derivative of both sides, and use implicit differentiation to solve for dy/dx:
d/dx ln(y) = r*d/dx ln(x)
1/y * dy/dx = r/x
dy/dx = r*y/x
Plug back in the original y:
dy/dx = r*x^r/x
Rewrite 1/x as x^(-1):
dy/dx = r*x^r * x^(-1)
Combine the exponent:
dy/dx = r*x^(r-1)
To prove that it ALSO works for the part of the function when x
@@carultch Thank you very much for this. It really helped me. Because till date I was proving the power rule only by the definition of derivative and binomial theorem.
Defining the derivative of ln(x) and exp(x) and using this was a nice idea. Thank you for your response.
The 2 explainations are very prompt. Thanks
You know, that’s impressive…
The differentiation rules are such direct corollaries of the definition, that it doesn't make sense to distinguish between them.
What is inside ur Pokemon ball 🤔 ?
His mic.
I was that student in my first go at college. After graduating with a non-mathy degree, I went back to school for CS.
"that student" will get by fine in a non-engineering degree. However, "That student" will flounder in an engineering degree program if they don't seek to understand the "why" offered by the long version.
At 5:43, instead of breaking it up like that, could you take out the (2^h - 1) term by itself, then find the limit of that (which is 0), then it would be the other limit times 0 which is 0? That's different than the answer, but I don't know where the mistake is with this way
(Im not 100% sure) he says that that splitting up of limits only works if both limits exist. Id assume the other limit doesnt as it still as the h in the denominator.