COOL Quotient Rule Proof
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- čas přidán 29. 08. 2024
- In this video, I give two super neat proofs of the quotient rule, one using multiplication, and the other one using logarithms. Enjoy!
Quotient Rule Proof: • Quotient Rule Proof
Quotient Rule in Greek: • Neat Quotient Rule Proof
Quotient Rule in Russian: • Cool Quotient Rule Proof
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I always hated that quotient rule and was always trying to finesse my way around it in my Calc I class. I used the product rule on
(f/g)' = (f * 1/g)' when I could.
Let's give that a 'try', again..
spoiler alert:
(f/g|)' = (f * 1/g)' = f'/g + f*(-1/g^2)*g' = (f'g-fg')/g^2
This, besides the product rule, also applies the -chain rule- Chen Lu, but does _not_ use implicit differentiation.
@@Apollorion Exactly how I've done it. Seems more straightforward and one less rule to remember.
Very cool video! When I was teaching Calc 3 last year, I learned that the multivariable chain rule can be used to prove the product rule and the quotient rule. Here's how it works with the product rule:
Define f(u,v) = uv. Then ∂f/∂u = v and ∂f/∂v = u.
If u = u(x) and v = v(x), then by multivariable chain rule
d/dx [f(u,v)] = (∂f/∂u)(du/dx) + (∂f/∂v)(dv/dx)
= vu' + uv'.
Quotient rule is the exact same thing but with f(u,v) = u/v.
That’s suuuuuper nice, thank you!!!
Never thought I'd subscribe to not just one but TWO math channels. This is a good medium to release and embrace my inner math nerd.
Nice!!! What’s the other one?
@@drpeyam Blackandredpen. Chen Lu.
Best channels ever!!!
In the second proof you can take Ln |f/g| = Ln |f| - Ln |g| and not have problems with signs. Remember that (Ln |x|)' = 1/x for both x0.
Oh yes, trueeee!! f might still be 0 though
@@drpeyam .. as can g
Edit: but of course the values of x for which g=0 are therefore not part of the domain of f/g, and henceforth also not part of the domain of (f/g)' so this does not matter.
No g can’t be 0 anyway, since otherwise f/g is undefined
It was very useful thanks sir
I like the second way of doing it because you don't even need the product rule for this proof while you do in the first one.
Loved the second method Dr Peyam!
Ah goodness logarithms are just inverses of exponentials but their algebra property laws are so very useful and beautiful!
This is awesome! I can't believe I made it through three years of calculus and real analysis and I never saw this!
I always hated the first derivation because part of the conclusion of the quotient rule is the existence of the derivative h=f/g. When you turn this into f=gh, you assume the existence of h'.
Now, as a memory trick, it is fine.
It’s so pretty though!!!
I used another method when I was young by using a small thing noted as Delta of x and Delta of y, which are small variations of x and y.
Could you please show your method? I wonder whether it brings something to the proof of if its equivalent to the known proofs.
@@PackSciences It would be difficult. I will try :
y = f/g
y + Dy = (f+Df)/(g+Dg)
Dy = (f+Df)/(g+Dg) - y
Dy = (f+Df)/(g+Dg) - f/g
Dy = (g(f+Df) - f(g+Dg))/g(g+Dg)
Dy = (gDf - fDg)/g(g+Dg)
Dy/Dx = (gDf/Dx - fDg/Dx)/g(g+Dg)
When Dx goes to 0, Df/Dx goes to f', Dg/Dx to g' and Dy/Dx to y' and Dg goes to 0.
y' = (gf' - fg')/g^2
@@alainrogez8485 Great
Hey Dr Peyam, thanks for sharing this video. I've been looking into your channel for 2 days now, and I've found that unlike most Math channels where the channel owners are Math PhDs/Teachers who end up doing videos about algebraic questions/ basic math stuff that simply aren't "interesting" (to me of course), you actually do COLLEGE level math in your channel, providing interesting insights in the world of high level mathematics. I've never found any other channel who does this in this high level, so you've earned my subscription, Keep going!
I was once taught the following:
(f/g)' = (f * g^(-1))'
= f' * g^(-1) + f * (g^(-1))' ( Product rule )
= f' * g^(-1) + f * (- g^(-2) * g') ( (1/g)' )
= (f' * g - f * g')/g^2 ( Common denominator )
I like your second proof a lot since it bypasses the product rule.
But I prefer this obe over the first rule since it briefly uses some simple rules.
This is really cool. I always forget whether it's g'f or f'g to start with.
Calculating with g^-1 is also an option, but this is nice, too.
A mnemotechnic is that the bottom part vehiculates the negative sign (g) in the final expression as (1/g)' would.
Hence, it is -g'f and not -f'g.
I always remember "Low d high minus high d low, square the bottom and away we go," as in, denominator times derivative of numerator minus the reverse
@@PackSciences sorry i left my multiphase quantum vehiculator at home
For number 1, I'm pretty sure you can do it in reverse, proving the product rule with the quotient rule. Just do the same steps in reverse.
I was going to go back and edit that post, but since it got a like, I won't. But I did finally have the time to sit down and prove the product rule given the quotient rule, thus showing the product rule is NOT more powerful than the quotient rule. Here is my proof:
Let f and g be differentiable functions, and let h = fg. Thus f = h/g, and f' = (h/g)'. By the quotient rule (which is given), f' = (h'g - g'h)/(g^2). By substitution, we get:
f' = (h'g - g'fg)/(g^2)
f' = g(h' - g'f)/(g^2) (factor)
f' = (h' - g'f)/g (simplification)
f'g = h' - g'f (division on both sides)
h' = f'g + g'f (addition on both sides)
Then, by substitution once again, we get
(fg)' = f'g + g'f, which is our product rule.
@ZipplyZane By assuming the quotient rule as true the argument kinda falls apart.
Let's say that "P" is the statement "The product rule is true" and "Q" is "The quotient rule is true" (both including that the functions to which they're applied are differentiable).
By stating that "the quotient rule is given" (Q is true by assumption) you only managed to prove that "if Q is true, P must also be true", or "Q→P".
The video shows a proof for "P→Q" (which is indeed the same as yours, but in reverse), so we have (P→Q)&(Q→P), otherwise written as "P iff Q".
What is needed now is a proof for either of them that doesn't rely on the other one being true (which would just be equivalent to P→Q or Q→P, which we've already established).
The point is that you can prove the product rule without relying on the quotient rule, and - as far as I know - you can't do it the other way around.
Stating that the product rule is a "stronger" theorem just means that it's the one holding up both of them.
Another way of saying this is: you can only come at the conclusion that Q is true only by first saying that P is.
I like imagining logical implications as "roads" that get you from a "city" to another. In this sense, you can't travel to Q without passing by P, thus making the road that connects them the only one leading to Q. P, on the other hand, has many more roads coming and going from it, like some sort of metropolitan city. In other words, P is more general than Q.
If you can find a proof for Q that doesn't need proving P first, you'll be right to say that neither of them is stronger since both of them have proofs indipendent from eachother's.
The same holds for the chain rule.
Chen Lu and Caution Lu 🙂
We just had to prooft the quotient rule in calc 2 and it was explicitly stated that the first method you showed was invalid since you are assuming that h’(x) exists which you are trying to show (same goes for the second although such an approach never came up)
lol, they must be fun at parties
If I understand:
The classic proof states that IF f', g', exist AND g != 0 THEN (i) (f/g)' exists AND (ii) (f/g)' =...]
Our first proof is valid, but moves the thesis (i) among the hypothesis.
[ I'm the serious lonely guy at the parties 🤣]
Easier? Let f and 1/g be functions. Use the product rule and the chen lu.
Nah, that is cheating, I feel 1/g is kinda like f/g anyway
Respect!
Use The Chen Lu!!!
How about :(f/g)’=(f*g-1)’= f’*g-1 + f(-g-2g’)=(f’g-fig’)/g2. Just straight application of the product rule.
This is nice - 2 different ways. Whenever I forget what the quotient rule, I turn it into a product like you did. I am curious -- Are you saying product "loop" and chain "loop"?
Product Lu and Chen Lu :)
@@drpeyam Thanks :)
Applying the ln, implicit differentiation and the chen lu can also proof us another very generally applied differentiation rule:
ln(x^u)=u*ln(x) => (ln(x^u))'=u/x=(1/(x^u))*(x^u)' (x^u)'=ux^(u-1)
Seems familiar?
Isn’t it just a product rule ? F’g-1+g-1’/f? :D
Smash the like button...This is so cool...yeah......
The ones I'm wanting to eventually do are to prove the product rule and/or quotient rule from the definition of the derivative. I've still not quite gotten it. But I hope to figure it out without being told. I did prove the power rule myself without help.
I'm also curious if there's a way to prove the ~chain rule~ chen lu from the definition of the derivative. I have an idea of how to do it with implicit differentiation, but the only "proof" I know of that uses a dummy variable and cancelling (i.e. I use dy/dt and dx/dt and multiply both by dt to get rid of the denominator).
There are videos about this on my channel
The sign error rule.
HERMOSO
It is called the qu shing lu
Dr Peyam is more brilliant than Gauss.
Dr. I am PI: How cool is that?
Me: it's super cool.
We call that consistency. This is what Godel's second theorem is all about.
If you really want to go hardcore Chen Lu, view f(x)/g(x) = u(f(x), g(x)) where u(x,y) = x/y. Then the multivariable chain rule gives the derivative as u_x ( f(x), g(x)) * f'(x) + u_y(f(x),g(x)) * g'(x) = f'(x)/g(x) - (f(x) / g(x)^2) g'(x) = (g(x)f'(x) - f(x)g'(x)) / (g(x))^2
That’ll be part of my future video
lou > rule
Less goooo
QUOTIENT LUUU!!!!! YEAH!!!!
You can cancle f one step before.
Why are you guys so mean to quotient rule?