Proving the power, product and quotient rules by using logarithmic differentiation & the chain rule

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  • เผยแพร่เมื่อ 23 ธ.ค. 2024

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  • @trucid2
    @trucid2 7 ปีที่แล้ว +173

    Implicit differentiation is like a magic trick. It feels like we haven't done any real work and still got the answer.

    • @Metalhammer1993
      @Metalhammer1993 7 ปีที่แล้ว +7

      well implicit differentiation just feels like a half assed chain rule to me. half assed because we don´t do half the work right away^^ (other than that it is a chain rule we could solve for y in a sepperate line and take the derivative but that would defeat the purpose of implicit differentiation. you only do half the work because you hope you can save quite a chunk of the work you just have ahead of you. like: figuring out the "inner function" algebraically)

    • @imperson1785
      @imperson1785 3 ปีที่แล้ว +3

      @@Metalhammer1993 it's also half assed because the domain can only be _half_ of all real numbers, i.e. x>0, since that's the domain for ln(x)

    • @williamestey7294
      @williamestey7294 3 ปีที่แล้ว +1

      Yes, all of these proofs only verify for positive real values of x. Proof of all real values requires more work.

    • @trucid2
      @trucid2 3 ปีที่แล้ว +1

      @@williamestey7294 Probably works on negative numbers too if you allow the result to be complex valued.

    • @williamestey7294
      @williamestey7294 3 ปีที่แล้ว +1

      Yes, that is correct. When using the generalized logarithm function defined for complex numbers, this proof can be shown to work over the entire complex plane.

  • @__nog642
    @__nog642 7 ปีที่แล้ว +245

    Damn, that power rule proof was way easier than the proof with the binomial theorem and the definition of the derivative.

    • @willsunnn
      @willsunnn 6 ปีที่แล้ว +34

      Neil Gupta , and that proof that uses the binomial theorem only proves it is true for positive integers

    • @KnakuanaRka
      @KnakuanaRka 6 ปีที่แล้ว +9

      Meh; I still find the old binomial version more memorable. Maybe it’s to do with it not requiring anything other than the basic definition of a derivative; no trickery or complications needed, just straightforward variable-juggling.
      And you really only need the first two terms of the expansion, since when you expand (x+h)^n, you can factor h^2 out of all the other terms, meaning they still have a factor of h after diving by h, so you know they all will go to 0 with the limit and can toss them out without having to write all of them; it’s not as messy or scary as it may seem at first blush.
      But perhaps it’s mostly because of a really cool geometric intuition shown by 3blue1brown in his Essence of Calculus series: th-cam.com/video/S0_qX4VJhMQ/w-d-xo.html That analogy seems to make it easy for me to get a grasp on the binomial version, while this one doesn’t have such an easy representation. Heck, he offers a lot of geometric representations of other derivative properties which are way more intuitive and easy to understand than any way of juggling variables.
      Then again, the binomial version only works for counting numbers, so I suppose we need this one anyways.

    • @yoavmor9002
      @yoavmor9002 5 ปีที่แล้ว +6

      @@willsunnn Well, this method only proves this within a function's positive domain. So theoretically, it shouldn't have to work with stuff like -x²-1 at all. You would need to add a limit as c approaches infinity of y+c to the left side to be able to use these formulas for every function, but that ruins the log properties

    • @blockthrower3947
      @blockthrower3947 4 ปีที่แล้ว +3

      I don't know if this is necessarily easier as you use one rule to prove the other rule rather than the binomial proof not relying on a more advanced derivative rule. And you usually learn the derivative of ln(f(x)) way after you learn the power rule.

    • @anshumanagrawal346
      @anshumanagrawal346 3 ปีที่แล้ว

      @@yoavmor9002 This was my first thought, seeing this proof

  • @harleyspeedthrust4013
    @harleyspeedthrust4013 3 ปีที่แล้ว +92

    i dont think i've used the quotient rule since high school calculus because it's usually easier to use the product rule with the denominator raised to the -1 power. this was a cool video though, i loved these derivations

    • @Ayush-yj5qv
      @Ayush-yj5qv 3 ปีที่แล้ว +3

      Same bro

    • @deniseeleam2320
      @deniseeleam2320 2 ปีที่แล้ว +7

      I am disappointed I’ve never thought of this but thanks for the tip

    • @fahrenheit2101
      @fahrenheit2101 2 ปีที่แล้ว +1

      I always used this to prove the quotient rule, but I never considered using it as an actual method

    • @harleyspeedthrust4013
      @harleyspeedthrust4013 2 ปีที่แล้ว

      @@fahrenheit2101 it works well. One less rule to remember and potentially mess up

    • @ThePeterDislikeShow
      @ThePeterDislikeShow ปีที่แล้ว

      I discovered that trick quite by mistake in high school because I thought that it was weird there are two different ways to look at a quotient and wanted to be sure they gave the same answer!

  • @ozzyfromspace
    @ozzyfromspace 5 ปีที่แล้ว +3

    My calc professor was Chinese and when teaching the quotient rule, she would always say, "low-d-high, high-d-low" with a Cantonese accent. I never forgot, she was an awesome instructor!

    • @tcmxiyw
      @tcmxiyw 2 ปีที่แล้ว

      One of my students remembered the quotient rule as “down d-up - up d-down all over down squared”

  • @angelmendez-rivera351
    @angelmendez-rivera351 6 ปีที่แล้ว +7

    Also, with regards to the power-to-the-power rule, if you distribute, you will find that dy/dx = g*f^(g - 1)*f’ + ln f*f^g*g’. This is interesting because it is the sum of the ordinary power rule as applied to y if g is treated as constant and of the exponential rule as applied if f is treated as a constant.

  • @brucefrizzell4221
    @brucefrizzell4221 6 ปีที่แล้ว +26

    proofs using ln() are very helpful. a lot easier to understand than the normal limit proofs.
    perhaps Blackpenredpen will explain Taylor polynomials.

  • @cpotisch
    @cpotisch 4 ปีที่แล้ว +8

    It’s amazing that with just the chain rule and the derivatives of ln and of sin, one can differentiate almost every single function.

    • @kyloben4848
      @kyloben4848 2 ปีที่แล้ว +1

      You only need e to the x to find the derivative of ln

  • @angelmendez-rivera351
    @angelmendez-rivera351 6 ปีที่แล้ว

    For those concerned with the commentary that this only works if x^r is positive, or if f*g or f/g or f^g is positive. This is not true. There is no restriction on what f, g, and x^r can be valued as long as the input x is real-valued. The reason is that the natural logarithm function as well as the exponential function are both well-defined if their domains are the set of complex numbers. Complex-valued functions are allowed inside the limit definition of derivatives, because the limiting variable and the inputs of those functions in the limits are real-valued, which is what is required. So f, g, and x^r can be complex-valued, with r being complex-valued, as long as x is real-valued. Hence there is no fallacy in what BPRP has done in the video.

  • @DaanSnqn
    @DaanSnqn 7 ปีที่แล้ว +46

    Fancy jacket 👌👌

  • @bmw123ck
    @bmw123ck 7 ปีที่แล้ว +17

    Could you prove the same but using the definition of derivative?
    Love your vids!!

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว +12

      bmw123ck yes. But just longer.

    • @bmw123ck
      @bmw123ck 7 ปีที่แล้ว +3

      Yeees, but not so obvious and systematic as it is using logarithmic differentiation

  • @pacolibre5411
    @pacolibre5411 3 ปีที่แล้ว +8

    Just realized something. The derivative of f^g is the sum of the derivative assuming g is constant (g*f^(g-1)*f’) and the derivative assuming f is a constant (f^g*ln(f)*g’)

    • @jesusandrade1378
      @jesusandrade1378 2 ปีที่แล้ว +1

      That is the verbal way of remembering the derivative of a function raised to another function, as shown in calculus books.

  • @mrteddy808
    @mrteddy808 7 ปีที่แล้ว +17

    Beautiful. Always wanted to know the proof, shame they don't show it in math B.

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว +4

      Teddy S what math B?

    • @timkw
      @timkw 2 ปีที่แล้ว +2

      @@blackpenredpen I dont know if this person is Dutch, but in the Netherlands we have different math classes: A,B,C,D. A&C are focussed on Statistics and Probability, with algebra and simple differentiation. B is focussed on Calculus, Trigonometry and Geometry. D goes further than A,B and C with complex numbers, group theory and matrices and some other stuff.

  • @gnikola2013
    @gnikola2013 7 ปีที่แล้ว +2

    In the case of the f(x)^g(x) derivative, you could write the final expression in a nicer way. If you distribute the f^g, you get f^g(gf'/f)+f^g(ln f) which is equal to gf^(g-1)f'+ln(f^g'). I find this expression nicer because the first term resembles a simple power rule and the second term is similar to the original function. Also, the first term doesn't include the g' and the second term doesn't include f', so the first one gives us information about f' and the second one about g'. I love this

    • @TDRinfinity
      @TDRinfinity 4 ปีที่แล้ว

      (x^a)'=ax^(a-1). (a^y)'=ln(a)*a^y. (x^y)'=yx^(y-1)*x'+ln(x)*x^y*y'.

    • @TDRinfinity
      @TDRinfinity 4 ปีที่แล้ว

      This is like a gradient or something like that

  • @General12th
    @General12th 7 ปีที่แล้ว +14

    _One dTwo plus Two dOne, that's the why we'll have some fun_
    _Low dHigh minus High dLow, square the bottom and away we go_
    Also, the integral by parts is _ultraviolet supervoodoo._

    • @KnakuanaRka
      @KnakuanaRka 6 ปีที่แล้ว +2

      J.J. Shank Interesting, though I generally remember which functions go where in the quotient rule by doing a quick sanity check with x/1. And I never bothered to use a mnemonic for integration by parts, both because of the D-I method used by BPRP and elsewhere, but before even that, it’s pretty trivial to get it from the product rule.

  • @ozzyfromspace
    @ozzyfromspace 5 ปีที่แล้ว +1

    Proof by logs is so easy it shouldn't be allowed lol.... technically these proofs constrain the functions (monotonically increasing, nonzero, etc) because logs blow up if you're not careful. Logs are great for sketching a proof before doing the heavy lifting. Fabulous video!

  • @JakeWaas
    @JakeWaas 6 ปีที่แล้ว +13

    Encore!
    Next, the chain rule please :D

    • @paulchapman8023
      @paulchapman8023 4 ปีที่แล้ว +7

      You can’t really use ln(y) to prove the chain rule, since its derivative in terms of x depends on the chain rule; it would be circular reasoning.
      It’s reasonably simple anyway: Let y = f(g(x)) and let u = g(x).
      By implicit differentiation, dy/dx = dy/du * du/dx.
      dy/du = f’(u) = f’(g(x))
      du/dx = g’(x)
      Thus dy/dx = f’(g(x)) * g’(x)

    • @rainbobow_8125
      @rainbobow_8125 3 ปีที่แล้ว +1

      @@paulchapman8023 hello man, I was wondering, is the substitution with u and y necessary or is it just to write it in a cleaner way ? Trying to understand without much knowledge sorry

    • @johnspence8141
      @johnspence8141 3 ปีที่แล้ว

      @@rainbobow_8125 chain rule is easy to prove using the product rule. Skip to about 4 min in: th-cam.com/video/5RyEpMHGg1Q/w-d-xo.html

  • @adblockturnedoff4515
    @adblockturnedoff4515 7 ปีที่แล้ว +3

    Interestingly enough if you do d(x^x)/dx you get (x^x)(1 + ln(x)). This was the same thing discussed in one of the previous videos using a different method to explain something different but I believe he got to the same result which I think ties up stuff nicely.

    • @harleyspeedthrust4013
      @harleyspeedthrust4013 3 ปีที่แล้ว +4

      i've heard that called the "generalized power rule." i remember looking it up a few years back for some CAS software because i had never learned how to differentiate a function raised to the power of another function. it's really nice to see where it comes from

  • @laugernberg4817
    @laugernberg4817 7 ปีที่แล้ว

    fun exercise: try to differentiate e^(kx), a^x, x^n etc. using the last formula at 17:40 :)

  • @thegoatman22
    @thegoatman22 7 ปีที่แล้ว +62

    ISN'T IT

    • @PubicGore
      @PubicGore 4 ปีที่แล้ว +1

      NAMELY

  • @borismezhibovskiy7607
    @borismezhibovskiy7607 7 ปีที่แล้ว +2

    Proving the power rule using e
    We start with some algebra to get it into the right form:
    1) y = x^r
    2) ln(y) = ln(x^r)
    3) ln(y) = r*ln(x)
    3) e^(ln(y)) = e^(r*ln(x))
    4) y = e^(r*ln(x))
    And now we're ready for the only piece of calculus in this proof, using the e rule and the chain rule:
    5) dy/dx = e^(r*ln(x)) * r/x
    Some more algebra to finish it up:
    6) dy/dx = e^(ln(x^r)) * r/x
    7) dy/dx = x^r * r/x
    And finally we arrive at the answer:
    8) dy/dx = r * x^(r-1)
    You're welcome ;)

    • @borismezhibovskiy7607
      @borismezhibovskiy7607 7 ปีที่แล้ว

      Similar thing can be done for the f^g rule. Why use implicit differentiation when the e method works?

    • @angelmendez-rivera351
      @angelmendez-rivera351 6 ปีที่แล้ว

      Steps 2, 3, and 3 (you wrote it twice) were unnecessary

  • @Supercat8835
    @Supercat8835 7 ปีที่แล้ว +5

    I really enjoyed this. Do you have any plans to prove the chain rule in a future video?

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว +6

      Patrick Conan I asked peyam for it already. Maybe we will work something out

    • @lukapopovic5802
      @lukapopovic5802 7 ปีที่แล้ว

      3blue1brown

  • @stephenbeck7222
    @stephenbeck7222 7 ปีที่แล้ว

    Of course the power rule in particular is fairly easy to show just with the limit definition of the derivative and some intuition about binomial expansion (the x^r term will cancel, and every term with h to a power greater than 1 in the original expansion will go to zero when you take the limit, so only one term remains) but using logs and implicit differentiation is a nice trick once you go through all the work of finding the derivative of the log.

    • @turbopotato4575
      @turbopotato4575 7 ปีที่แล้ว +1

      the binomial proof is only valid for integer powers this is valid for all

  • @12-343
    @12-343 3 ปีที่แล้ว +2

    I thought I'd share the mnemonics my first calculus teacher gave us for the product and quoptient rules:
    1 d 2 plus 2 d 1, calculus is so much fun
    and
    2 d 1 minus 1 d 2, draw a line, and square below

  • @dancapps3374
    @dancapps3374 2 ปีที่แล้ว +1

    Why is the derivative of ln(x) video private?

  • @LouisEmery
    @LouisEmery 2 ปีที่แล้ว

    Nice. One can check with either f or g as a constant or x and verify that it all works out.

  • @johnspence8141
    @johnspence8141 3 ปีที่แล้ว +1

    The easiest way to prove the quotient rule is the use the product rule and make the denominator function to the power of -1. Proceed with the product rule and use the power rule for the -1

  • @atomix1093
    @atomix1093 7 ปีที่แล้ว +32

    I've got an interesting question.
    When you have (-1)ⁿ in a series, it means that the signs of terms will be alternating (+-+-). Does there exist a similar formula that would make every 3rd (or nth for that matter) term have a different sign (++-++-)? Or maybe even follow some kind of pattern, like 2 positive and 3 negative, etc?

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว +37

      AtomiX omg! That's actually on my to-do list. I used to ask my students about finding a formula for 0,0,1.
      Hint: use a trig function

    • @atomix1093
      @atomix1093 7 ปีที่แล้ว +4

      blackpenredpen Thanks, I'll look into it :)

    • @MrSnowy737
      @MrSnowy737 7 ปีที่แล้ว +4

      You could also use a rounding function like the floor function, eg (-1)^Floor[n/2] or a combination of them to make a pattern

    • @z01t4n
      @z01t4n 7 ปีที่แล้ว +7

      AtomiX It's easy to generate patterns like this using sine/cosine. For example, "1-sqrt((sin(n*π/3))^2)*2/sqrt(3)" is equal to 0 at n=1, 2, 4, 5, 7, 8; and equal to 1 at n=0, 3, 6, 9. Of course, you can do it with -1 and 1 as well by using different constants.

    • @atomix1093
      @atomix1093 7 ปีที่แล้ว +4

      I see, it's not as complicated as I thought it would be

  • @15schaa
    @15schaa 7 ปีที่แล้ว +2

    What's a mathetician's favourite talk show host?
    ln(DeGeneres).

  • @Bhamilton-ws4go
    @Bhamilton-ws4go 6 ปีที่แล้ว +1

    Very simple and elegant!

  • @andrewcook1428
    @andrewcook1428 6 ปีที่แล้ว +3

    New favorite proof, law of cosines or geometric summation are close seconds

  • @ЮрійЯрош-г8ь
    @ЮрійЯрош-г8ь 6 ปีที่แล้ว +11

    Can we really take natural logarithm on both sides ? x^r can take negative values so we probably can't take logarithm ?

    • @xyzzyx9431
      @xyzzyx9431 6 ปีที่แล้ว +4

      Юрій Ярош
      I think you're right, you cannot take the logarithm if x^r is negative. But there is a simple way to proof that the formula is also right for negstive values of x^r. Let's say the formula in the video is right for all x>0 in case r is odd (so x^r for x

    • @xyzzyx9431
      @xyzzyx9431 6 ปีที่แล้ว +2

      Юрій Ярош
      A similar method can be used for the other proofs, I'm just gonna show it on the product rule for the others it works the same way. If y

    • @happygimp0
      @happygimp0 5 ปีที่แล้ว +4

      What stops you from taking the ln of a negative number?
      ln(-R) = i*π + ln(R) with R ∈ ℝ ∖ 0 (I am not sure what happens when R is not real)

    • @cerwe8861
      @cerwe8861 4 ปีที่แล้ว +2

      @@happygimp0 and if we differentiate, the imaginary part becomes 0, so we doesn't have to care about neg. Numbers.

  • @dqrksun
    @dqrksun 3 ปีที่แล้ว +1

    The last one should be called the REAL power rule

  • @hydropage2855
    @hydropage2855 3 ปีที่แล้ว +2

    Looking sharp, professor Chow hahaha

  • @kevincaotong
    @kevincaotong 7 ปีที่แล้ว +2

    Hey Blackpenredpen. I love your videos about calculus and I am currently stuck on an integral. I believe it involves the Zeta Function, which I could not find anywhere relating to integrals and solving them. Can you do a video on this integral please: integral from 0 to infinity of x^7/(1-e^x) dx. Thank you!

  • @ilkinond
    @ilkinond 4 ปีที่แล้ว

    That was a really beautiful presentation.

  • @polyd2500
    @polyd2500 7 ปีที่แล้ว +2

    Thanks, this really helped with my year 5 maths exam!

  • @thomasblackwell9507
    @thomasblackwell9507 4 ปีที่แล้ว

    I really enjoy your proofs and derivations.

  • @yassinebouih2097
    @yassinebouih2097 2 หลายเดือนก่อน

    تبارك الرحمن، افضل قناة..

  • @thomasg6830
    @thomasg6830 7 ปีที่แล้ว +3

    So good. 😎
    And i like the outro music.

  • @EgoTeach
    @EgoTeach 2 ปีที่แล้ว

    LOVE YOUR VIDEOS MATE. I'm taking IB Math AA HL and damn your videos are fine.

  • @arts5852
    @arts5852 4 ปีที่แล้ว

    Very cool explanation. I did not know it before, but now I understand how to get these rules. Thank you 👍

  • @OonHan
    @OonHan 7 ปีที่แล้ว +1

    d/dx (x^x) thus is equal to x^x(ln x + 1)
    Thanks!!!!! I finally know!!!!!

  • @nuklearboysymbiote
    @nuklearboysymbiote 5 ปีที่แล้ว +1

    but u can only take natural log of positive numbers… so why does this apply to negative y?

    • @happygimp0
      @happygimp0 5 ปีที่แล้ว

      What stops you from taking the ln of a negative number?
      ln(-R) = i*π + ln(R) with R ∈ ℝ ∖ 0 (I am not sure what happens when R is not real)

  • @SicilianTM
    @SicilianTM 7 ปีที่แล้ว +1

    Product: One prime two plus two prime one, isn't mathematics fun?
    Quotient: Low d high less high d low, over the denominator squared we'll go.

  • @1anya7d
    @1anya7d 7 ปีที่แล้ว +3

    Came for the bonus, now back to my stupid test

  • @SlingerDomb
    @SlingerDomb 7 ปีที่แล้ว +1

    let me ask you a question "why you didn't put the abs sign when you take the ln both side?" is that because your assumption in the earlier? assume all function are nice. thanks

  • @muslimahtutoress266
    @muslimahtutoress266 3 ปีที่แล้ว

    This looks GREAT!! Thanks

  • @hgaclark9031
    @hgaclark9031 7 ปีที่แล้ว +4

    Hi , please can you help me with a question?? It is:
    What is the sum of the following terms:
    1+e^(-x)+e^(-2x)...
    And over what range of x is the solution valid?

    • @turbopotato4575
      @turbopotato4575 7 ปีที่แล้ว +1

      Its a simple geometric series
      1+e^(-x)+e^(-2x)... = 1+e^(-x)+(e^(-x))^2... = 1/(1-e^(-x))
      Its convergent when e^-x < 1 so x > 0

    • @kevincaotong
      @kevincaotong 7 ปีที่แล้ว +1

      If we use a substitution, u=e^x, we can simplify the sum as 1+ 1/u + 1/u^2 +... This an infinite geometric series, equal to 1/(1-1/u). Expanding u, we get 1/(1-e^-x). I'm guessing you mean over what domain of x does the solution converge, which is when |e^(-x)|0, which is the domain of the solution.

    • @ayushranjan6807
      @ayushranjan6807 7 ปีที่แล้ว +1

      The sum is a geometric progression (GP), with first term a=1 [basically e^(0)] and common multiple r=e^(-x).
      (HOW - 1st term is e^0, 2nd term is [e^(-x)], 3rd term is [e^(-x)]^2 = [e^(-2x)], and so on)
      Now general formula for sum of a GP is S = a[1-(r^n)]/[1-r], where n is no of terms in the progression.
      If n-->inf, then for the sum to be convergent (approach to a finite number), |r|

  • @ryanlian7656
    @ryanlian7656 7 ปีที่แล้ว +1

    Blackpenredpen, can you go over more about complexifying an integral? I'm currently in calc 2 and I'm loving it so far so I want to be very good at integrals, do you know where I can find hard practice problems with answers/go over process ?

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว

      Ryan Lian u can check out my playlist of integral battles. th-cam.com/play/PLj7p5OoL6vGzuQqjZfMsAX7-R5fNji0cO.html

  • @troybingham6426
    @troybingham6426 4 ปีที่แล้ว

    Wow that's good. I'd never seen those formulas proven that way before. .. Nice.

  • @davidrheault7896
    @davidrheault7896 7 ปีที่แล้ว +1

    Great video with wonderful explanations :)

  • @magefreak9356
    @magefreak9356 3 ปีที่แล้ว

    Love the enthusiasm ❤️

  • @NonTwinBrothers
    @NonTwinBrothers 4 ปีที่แล้ว +3

    I can't stop watching your videos!! aaaaaaaaaaagh

  • @July-gj1st
    @July-gj1st 6 ปีที่แล้ว +1

    Is it possible to do a similar thing for integration?

    • @Salarr
      @Salarr 4 ปีที่แล้ว +2

      integration by parts can easily be proven via product rule. not sure about integration by substitution but I imagine it is easy to prove via chain rule

  • @Alberto_4fun
    @Alberto_4fun 7 ปีที่แล้ว

    You also can prove the product rule with the binomial theorem, using that in the derivative limit definition, using (x+h)^n as the binomio. I'm from México, so, sorry about my english

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว

      Magaña Drums u mean the power rule. Also that only works if n is positive whole number

    • @Alberto_4fun
      @Alberto_4fun 7 ปีที่แล้ว +1

      Oh yes yes, the power rule, sorry

  • @GAMERCREEDbrickjet
    @GAMERCREEDbrickjet 7 ปีที่แล้ว

    for the last one, you could have arranged the bracket to show it was product rule for the function g*ln(f), and then written it as (g*lnf)'

  • @tanelkagan
    @tanelkagan 2 ปีที่แล้ว

    Could we have a little more insight into what is meant by a "nice" function? Does it mean bijective? Continuous? Differentiable? Or something else?

    • @carultch
      @carultch ปีที่แล้ว

      In this context, it means continuous, differentiable, with a single-valued output.

  • @gmo2119
    @gmo2119 3 ปีที่แล้ว

    This is a easy way to culculate derivative rules, but I think it will be not sufficiented for proofing.
    Because LN(x) is positive(>0) all of x. so if y(x) or f,g(x) is negative(

  • @chessandmathguy
    @chessandmathguy 7 ปีที่แล้ว

    neat and well presented. love it!

  • @DancingRain
    @DancingRain 7 ปีที่แล้ว +3

    There was a mnemonic a classmate taught me in college for the quotient rule: "Low D high minus high D low, square below"

  • @helloitsme7553
    @helloitsme7553 7 ปีที่แล้ว +1

    But isn't the derivative of ln(x) based on the derivative of e^x which is based on its expansion, which requires the power rule?

    • @MrSnowy737
      @MrSnowy737 7 ปีที่แล้ว +7

      HelloItsMe the derivative of e^x doesn't require the power rule, it only needs the definition of e and limits

    • @gaurav.raj.mishra
      @gaurav.raj.mishra 7 ปีที่แล้ว +1

      HelloItsMe Derivative of ln(x) can be proved without e^x.

    • @leonardromano1491
      @leonardromano1491 7 ปีที่แล้ว

      Lookup "Differential algebra". An exponential is a function where differentiating is the identity, e.g. dF=id(F) then F is an exponential. Since you only need the chain-rule to show what the derivative of the inverse function is, it is sufficient to have both this definition of an exponential and the chain-rule for the derivative of it's inverse function.

    • @ayushranjan6807
      @ayushranjan6807 7 ปีที่แล้ว +2

      The derivative of any function (if it exists ofc) can be obtained using "first principle". Knowledge of derivative of other functions isn't reqd for that, knowledge of limits is.

  • @trucid2
    @trucid2 7 ปีที่แล้ว

    Nice. So we can get the derivative of x^x using the general formula, giving us the same answer: x^x*(1+ln(x))

  • @AhmedHan
    @AhmedHan 3 ปีที่แล้ว +1

    You can't take ln() of every function.
    The domain matter my friend. You ignore the domain in a lot of videos.

  • @nicolasmalo8178
    @nicolasmalo8178 4 ปีที่แล้ว

    Wooow ! But what if g or f is negative ??

  • @comprehensiveboycomprehens8786
    @comprehensiveboycomprehens8786 7 ปีที่แล้ว +1

    This is just the innocent property of logarithms. :).

  • @nabeelkhan4910
    @nabeelkhan4910 6 ปีที่แล้ว

    U can also prove differentiation of constant power of x

  • @shivajigarg2383
    @shivajigarg2383 7 ปีที่แล้ว

    If we are taking the log of everything, is it not necessary to have all of them a positive number or positive function. So are these proof ligitimate.

    • @angelmendez-rivera351
      @angelmendez-rivera351 6 ปีที่แล้ว

      Shivaji Garg Not really. If we use complex numbers, then we are allowed to use any real function or number as input.

  • @jkstudyroom
    @jkstudyroom 4 ปีที่แล้ว +1

    Hey, you're using circular reasoning to find the derivative of natural log to prove the power rule.

  • @naimulhaq9626
    @naimulhaq9626 7 ปีที่แล้ว

    blackpenredpen
    Please do a video on differential logarithm, please.

  • @popalofiti480
    @popalofiti480 7 ปีที่แล้ว

    We know all of these derivatives, yet it feels one is quite elusive.
    What is the derivative of x factorial?

    • @alexander51413
      @alexander51413 7 ปีที่แล้ว +2

      f(x)=x! is not continuous, so it doesn't really have a definitive, but if do some fancy analytic continuation and a lot of maths, you can end up with something called the gamma function (look it up - en.wikipedia.org/wiki/Gamma_function), which is essentially the same thing (Gamma(n)=(n-1)!) and that does indeed have a derivative.

    • @farrela6710
      @farrela6710 4 ปีที่แล้ว

      search up the gamma function and the digamma function :D

  • @lucasnunes5185
    @lucasnunes5185 6 ปีที่แล้ว

    Gee, someone knows in which video his friend Payam makes a tower of e's?

  • @friedkeenan
    @friedkeenan 7 ปีที่แล้ว

    Why do people always say the power rule is only for real numbers? I saw no operation in this proof that was real number specific, so why can't it work with complex numbers?

    • @carultch
      @carultch ปีที่แล้ว

      It can and does work for complex numbers as well. People just say it is limited to real numbers, to keep it simple as a Calc 1 topic. There is a proof that it works for complex numbers as well, using the complex logarithm.

  • @isambo400
    @isambo400 5 ปีที่แล้ว

    What if I have no g of f?

  • @jarikosonen4079
    @jarikosonen4079 4 ปีที่แล้ว

    Differentiate the x^x with this power to the power rule... d/dx( x^x ) = x^x * ( 1 + ln(x) ) ?
    Maybe this all useful {f(x), g(x)} rules there is?

  • @paulh4828
    @paulh4828 4 ปีที่แล้ว

    0:53 And that works only for all r€lR, x>0. So you proved it for a certain
    f : lR*+ ---> lR
    x ---> x^r
    And you did not generalize it for
    g : lR ---> lR
    x ---> x^r

  • @xyBubu
    @xyBubu 4 ปีที่แล้ว

    the video about the derivative of ln(x) is private

  • @ayalacuetoricardoantonio5282
    @ayalacuetoricardoantonio5282 2 ปีที่แล้ว

    I solved my existential crisis

  • @igoralvespratescostaleite3577
    @igoralvespratescostaleite3577 4 ปีที่แล้ว

    Wonderful!

  • @aizek0827
    @aizek0827 4 ปีที่แล้ว

    We have to use the CHEN LU!

  • @dcauz
    @dcauz 2 ปีที่แล้ว

    I prefer dy/dx = f^(g-1)*(g*f'+g'*f*ln(f)) as the final form.

  • @arielfuxman8868
    @arielfuxman8868 4 ปีที่แล้ว

    You can prove the quotient rule using the power rule and the product rule, no need for ln

  • @jihanhamdan5465
    @jihanhamdan5465 7 ปีที่แล้ว +1

    Awesomeee i love logarithms

  • @sandmann6851
    @sandmann6851 7 ปีที่แล้ว

    Link doesnt work

  • @nabeelkhan4910
    @nabeelkhan4910 6 ปีที่แล้ว

    Dapper jacket!!! Boii!!

  • @PedroHenrique-zy3uh
    @PedroHenrique-zy3uh 7 ปีที่แล้ว

    Awesome!

  • @fernandobermudez2062
    @fernandobermudez2062 5 ปีที่แล้ว

    Easier way to prove the power rule is with mathematical induction

  • @john-athancrow4169
    @john-athancrow4169 6 ปีที่แล้ว

    Completion: The order of addition doesn't matter.

  • @mobilkonto__free__9726
    @mobilkonto__free__9726 6 ปีที่แล้ว

    Handsome af and smart

  • @Gilltes
    @Gilltes 7 ปีที่แล้ว

    Hello, can you try and integrate e^x*cos(x) without using integration by parts?

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว +1

      Mário Marques does complexifying the integral count? If so, I did that already.

    • @Gilltes
      @Gilltes 7 ปีที่แล้ว

      blackpenredpen You mean converting the expression in the integral to a complex expression? Yes, thats how I want to solve it.

    • @blackpenredpen
      @blackpenredpen  7 ปีที่แล้ว

      Mário Marques here th-cam.com/video/m0C70WQDHNg/w-d-xo.html

    • @Gilltes
      @Gilltes 7 ปีที่แล้ว

      blackpenredpen Amazing, thanks

  • @bluegrovyle5480
    @bluegrovyle5480 7 ปีที่แล้ว

    Mathematics is just the perfect example of the infinite power and immutability of God. Everything in math is circular reasoning, and we do things just because they always have and always will work. How much more trust, then, can we place in the God of the Bible who promises that all things will contribute to the good of those who love Him, who have been called according to His purpose?

    • @angelmendez-rivera351
      @angelmendez-rivera351 6 ปีที่แล้ว

      Blue Grovyle 1. Mathematics has no relation to any entities, let alone any deities of any religion such as the so-called God you talked about.
      2. No, mathematics is not circular reasoning. Mathematics constitutes a set of formal interpreted theories, which contain a set of chosen axioms, and every theorem is derivable from these axioms based on truth-preserving inference rules, which is determined by the semantics of the language on which the axioms are formulated.
      3. Yes, we do these things because they work, and as such, they are reliable. However, this does not warrant that we put any trust in deities, because it cannot be demonstrated that these deities are reliable, which is not the case with mathematics, for which you can prove its reliability. Therefore, under the burden of proof, we cannot assume God or the Bible are reliable, as they are not connected to mathematics - and in fact, statements in the Bible are often in direct contradiction with mathematics - and thus, we should not rely on them until proven otherwise.
      4. Can you provide evidence or proof that God exists? If not, can you then provide a reason why we should simply have faith instead of obeying the burden of proof, which is the only reliable epistemic principle by which we can operate thus far?

  • @ViralVision
    @ViralVision 7 ปีที่แล้ว +3

    Prove, using differential equations and laplace transforms, that sin(ix) = isinh(x)

  • @fountainovaphilosopher8112
    @fountainovaphilosopher8112 7 ปีที่แล้ว +1

    Cool.

  • @allaincumming6313
    @allaincumming6313 7 ปีที่แล้ว

    Logarithms rules!

  • @teodorlamort3864
    @teodorlamort3864 7 ปีที่แล้ว

    logarithms of negative values yayy

  • @blackholesun4942
    @blackholesun4942 9 หลายเดือนก่อน

    4:30

  • @nabeelkhan4910
    @nabeelkhan4910 6 ปีที่แล้ว

    Prove the chain rule!!!!!!

  • @mikejackson19828
    @mikejackson19828 3 ปีที่แล้ว

    Isn't it.

  • @dogbiscuituk
    @dogbiscuituk 7 ปีที่แล้ว

    Can these derivations be extended into the regions x < 0, and/or y < 0? - I mean, without complexification!

  • @casa1420
    @casa1420 6 ปีที่แล้ว

    Muito bom!

  • @johnjenuga3548
    @johnjenuga3548 7 ปีที่แล้ว

    nice