Which definition of derivative is easier?
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- Опубликовано: 1 окт 2024
- As we have seen in calculus 1, there are two versions of the definition of the derivative of a function at a, namely f'(a). BUT which one is easier? We will work out both and you will let me know which one you prefer! Subscribe for more calculus tutorials 👉 @bprpcalculusbasics
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I use them both depending on the need. Ex: solving functional eq. Given a differentiable f s.t. f(x+y)=f(x)f(y) you can use the second definition to get that f' = Cf which is a separable DE. But if functional eq. has some sort of a sum or a difference in it - then the first definition is better.
b is easier since in order to cancel with the denominator you just have to factor out a single h which isn't really hard, but in a) you have to factor out a difference which is harder.
b) is just algebra autopilot
Option b is the best one other than power rule
a) is easier acc to me..reason is that it feels quite simple to take factors rather than expand
In fact all are nice
I tend to prefer the h->0 version, I have a lot more experience with that one, and at least for polynomials, expanding is easier than factoring, but I know there are times when the x->a version is better.
Thank you very much for making these kinds of videos.
I would prefer version b)
I didn't even know about the pascals triangle thing. I hate the definition more than anything else in calculus. The easier of the two depends less on the need and more on your ability to factor versus multiply.
There is a really easy shortcut to expand binomials as you write them out.
Say you need to expand (A+B)^7. Start with A^7+7(A^6)(B^1). Then the cooeficient for the next term will be the coefficient of the most recent term (7) times the exponent on A (6) divided by ONE MORE than the exponent on B (1). So we have 7 * 6 / (1+1) = 21 and the expansion so far is A^7+7(A^6)B+21(A^5)(B^2). Just do the same thing to find the coeefentient of the next term. 21*5/(2+1). In this case 21/3=7 and 7*5=35 so we have A^7+(7A^6)B+21(A^5)(B^2)+35(A^4)(B^3). At this point we can simply mirror the rest yielding: A^7 + 7(A^6)B + 21(A^5)(B^2) + 35(A^4)(B^3) + 35(A^3)(B^4) + 21(A^2)(B^5) + 7A(B^6) + B^7.
Hopefully this makes sense.
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You didn't compare central differentiation. :)
I much prefer method 2. Not only is expanding usually easier than factoring (as you said, in method 1 you sometimes even need long division), but also it makes finding the derivative of sin(x), cos(x) and exp(x) easier, it is easier to prove the product rule and the chain rule using that method, and it is easier to write down linear approximations (tangents), which you both need for deriving Newton's method for finding roots and l'Hospital's rule.
So used to seeing the 2nd definition of derivative from FTC that I forgot about the first one! Haven't seen it for a while. I can see advantages for both (h always approaching 0+, while x always approaching a)
I actually like both, they both have their own implications
Although I am more familiar with b i like a more
I’m a B kind of guy
Preference: Proofing the power rule using implicit differentiation. ( *or method 2* )
Before that, we'll proof the derivative of ln x using the second definition of the first principle mentioned. Then there is no need to do that awful binomial expansion.
Or just recall that all terms with degrees greater than h^2 would be insignificant, and thus we could just ignore those terms in each binomial and do not expand them fully. (by using the second definition of the first principle mentioned)
(a+h)^99 = a^99 + (99C1) a^98 h + (insignificant terms containing h of degree 2 or above)
= a^99 + 99 a^98 h
((a+h)^99-a^99)/h = (99 a^98 h + insignificant terms)/h = 99 a^98 + insignificant terms that contains h.
P.S. if you would like a more formal and mathematical formulation, then just keep all insignificant terms in an summation of binomial expansion under the summation notation. Calculating their exact coefficients is just not necessary.
Neither one, of course as a calculus student who already learned derivatives I would say….
The epsilon-delta definition of either limit
Although I have just started learning calculus I prefer the 2nd method because it just works better for me. Thanks for these vids bprp!
i luv maths
I prefer option b just for clarity reasons, because the increment h on which the whole idea of derivative is based is actually clear immediately. Although sometimes I might just substitute h with x-a and then solve the limit since the first and the second option are simply related by a sub
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prefer part b. I cannot explain why?
a. is easier to calculate 0/0 limits
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i love calculus
They're the same methods.
Can you find the derivative of e^x with the first definition?
Yes - but your def of the natural log changes
I don't know or study anything about calculus, but I love videos
I disagree. The first one you used a difference of cubes to factorise, and you couldve done the same with the second one. If you do that then honestly theres not much difference
“Innocent 2”
Part b
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Why lim_h>0: [f(a+h)-f(a-h)]/(2h) is not used more frequently as a definition of "derivative", even if the function has a removable discontinuity at a?
I think it's because in order to build a good differentiation theory, everybody agrees that differentiability must imply continuity, so I guess that's why
i always use the second one
In my opinion, both are the same, but the left form it's easier than the right form, why? The first one always is factorization (although both have this, however, it's "easy" to find out in this form). On another side, if we have a polynomial equation of 3rd or 4th grade, and evaluate at f(x+h) it's a little bit complicated to work with that, therefore, in my opinion (again) the first one is easier.
"it's "easy" to find out in this form"
Most of my students would beg to differ. ;)
I promise you. Your opinion counts.
a