Logged in to say thank you very much! I'm in intro to Real Analysis and was struggling to understand this proof in my textbook (I also noticed there were a couple typos which was really throwing me off). You did a great job of explaining this proof, and I was able to completely rewrite it from memory after having you explain it! I'm in school right now to be a high school math teacher, and I hope to explain mathematics as well as you in a couple years. Thank you!!!
G'(0) = 1 is correct. Yep, it's typically these unusual functions that we wouldn't normally think of right away that show us these strange properties, or that show us we can't assume things that seem intuitive, etc.
Changing my previous answer.. If g(x+h)=g(x) for only finitely many values of h near 0, then we can eventually get h "close enough" to 0 so that g(x+h) doesn't equal g(x). In other words, we can find a small enough open interval centered around x such that g(a) and g(b) are different for every a and b in the interval. This means g must be strictly increasing or decreasing on that interval. I think the intermediate value theorem can be used to formally prove that. Since we take the limit as
It's good to ask these questions. The derivative of f(g(x)) still exists regardless of what F is. But F has to be chosen in such a way that we can use it to actually show the derivative exists. If we choose F(y) to be constant when y = g(x), then this F will not work. That doesn't actually mean the derivative of f(g(x)) doesn't exist, it just means that the chosen F will not help us show the derivative exists. Retaining that equality you mentioned earlier is important, but we need more
Good question. Short answer: If you define F(y) to be a constant when y = g(x), then F(y) is not guaranteed to be continuous. But F(y) has to be continuous for this proof to work. More details: In the video between near 15:40 and near 19:40, I show that the limit as t approaches x on the first piece of F(g(t)) is equal to the limit as t approaches x on the second piece of F(g(t)). This equality shows that F(g(t)) is continuous, and F(g(t)) must be continuous or else the conclusion stated
On the surface it feels like there's too much of a dependency between x, r, and z. For the proof to be valid we need x and z to be arbitrary, but z depends on x and r since x < z < r. Making r arbitrary would also be a problem since there's a dependency between x and r. But I can't come up with any examples where the proof fails. It might require going beyond polynomials.
Why is it that f(g(t))-f(g(x))/g(t)-g(x)=g(t)-g(x)/t-x? That part of the proof (around 18:45) felt kinda like you were just saying they looked similar and therefore they're identical. However, if f(x) is 3x and g(x)=2x, for instance, then 6t-6x/2t-2x is not equal to 2t-2x/t-x, is it? One simplfies to 6(t-x)/2(t-x)=3, and the other simplifies to 2(t-x)/t-x=2. So, from this definition, I've derived that 3=2, which is a contradiction. This statement would only hold true if f(x) were equal to g(x), which only proves the chain rule for f(f(x)). Where is my mistake?
Caleb Spiess What did I say/do around 18:45 that makes you think I suggested those two expressions are equal? I watched the video around that time and didn't see any indication of that.
Oh, after watching it like the fifth and sixth times over I realized what you meant lol. f(g(t))-f(g(x))/g(t)-g(x) is literally the limit definition of f'(g(x)). I don't know why I got so confused at that part, but now I understand the whole proof. Thanks, by the way! The proof on khanacademy is "start with dy/dx and multiply by du/du to get dy/dx=dy/du * du/dx" which was completely unconvincing to me, so this video helped me get a feel for why the chain rule REALLY works.
Is the plan to use the CMVT to say that [f(r) - f(x)]/[g(r) - g(x)] = f'(z)/g'(z), and then apply the hypothesis? If that's the case then we would always need x < z < r, and I'm not entirely sure we can always have that if we also want abs( [f(r) - f(x)]/[g(r) - g(x)] - f(x)/g(x) ) to be less than e/2. For that inequality, we would need x and r to be "close enough" to each other. It may be the case that getting x and r "close enough" might not be possible if we also must have x < z < r.
Do you mean at a specific value of x, or for any value of x? For a specific x, not necessarily. Consider a variation of the function I mentioned in another response: G(x) = x^3 * sin(1/x) + x if x is not 0, and G(0) = 0. This function is continuous and differentiable at x=0, and it's true that G(0+h)-G(0)=0 for infinitely many values of h near 0, but G'(0) is not 0. For any x in general, I think the conclusion would still be not necessarily for similar reasons.
I think such a function would have to be oscillating, like G(x) = x^3 * sin(1/x) when x is not 0, and G(0) = 0. This function is continuous and differentiable at x=0. Also, this function oscillates near x=0, and it oscillates so much that no matter how close h gets to 0, there is always "another" h closer to x=0 so that G(0+h) = G(0).
than that. We need an F that both retains that equality and is continuous. The F chosen in the video meets these requirements. If we choose F to be constant when y = g(x), then F retains the equality but is not continuous. Since F is not continuous, then the last step in the proof becomes invalid. And to reiterate the earlier point, that doesn't mean the derivative of f(g(x)) doesn't exist, it just means that the F we chose won't help us find the derivative.
Thanks for this nice educative lecture. Though my question is about how you have defined F(y). You have given condition for g(t) and g(x) but you have not given us the conditions of t and x in this piecewise function, that is for the upper and lower expressions. Can you please clarify on this. Thanks in advance
But even if we could conclude that g'(x) = 0, dividing by g(x+h) - g(x) would still be a problem because 0 * (number) = 0 only if (number) is some finite quantity. If g(x+h) - g(x) is 0 for infinitely many h near 0, then {f[g(x+h)-f[g(x)]}/{g(x+h)-g(x)} is undefined for infinitely many h near 0, and undefined * 0 is still undefined. It'd basically be 0/0, which is one of the indeterminate forms. I'm not sure if that's where you were going with that but I wanted to mention it to be thorough.
You say that we don't know what g does, but you seem to assume that g(t) -> g(x) when t -> x. Is this a property of differentiable functions or am I missing something?
Gliffie Sounds like you got it now, but in case anyone else comes here later wondering the same thing, the fact that g(t) -> g(x) when t -> x follows directly from the fact that g is a continuous function (and can be proven using epsilon-delta techniques), and g being continuous follows directly from the fact that g is differentiable.
Took me a while but i finally got there i think! thanks! I have one question though, at 8:36 you say that the g(t) - g(x) cancel out but at the beginning at 1:54 you say they cant cancel out because of infinitesimals. surely infinitesimals would apply here too then? I wld ask you about the product rule at 14:!4 but ur already answering that on another video ahahahah thanks dude
More details: Showing something like this doesn't exist in general is much harder than showing that it does exist. In this case, to show that the derivative of f(g(x)) exists, we only need to pick one function F that works. If we pick one function F that doesn't work, we can't conclude anything yet. If we wanted to go this route to show the derivative of f(g(x)) doesn't exist, we would have to show that the proof wouldn't work for EVERY function F we could choose.
h goes to 0, we only care what happens when h is "very close" to 0. If we can make h "close enough" to 0 so that g is strictly increasing or decreasing on a tiny interval around x [the interval would be (x-h,x+h)], then yes, we can use the "flawed" proof. The problem is that f and g are arbitrary, so we can't guarantee g has that property. If g doesn't have that property then we can't use the flawed proof. But I'm having trouble thinking of a (nontrivial) function without that property.
Why do f and g go to 0 as x goes to infinity? That doesn't hold for any nontrivial polynomial or for many types of transcendental functions. If we begin with that as an assumption then it definitely looks like this all works out, but it's a pretty restrictive hypothesis to have.
Hi! I would like to add that f(g(x)), its derivative, is not df/dx. And it's not df/dx= df/dg * dg/dx. But instead it should have been y=f(g(x)) and dy/dx=dy/dg*dg/dx which is equivalent to [f o g]'(x)= f'(g(x))*g'(x). And the dg'es do cancel each other since they are actually just a number (an extremely small one) with limit -> 0 (infinitesimal). Is that correct sir? thanks!
Euvlid W your first comment is right, my mistake. Your second comment is not correct. Infinitesimals are not real numbers (like 1, 3.7, -12, pi, sqrt(5), etc. are) and cannot be treated as numbers, just like how infinity is not a number and cannot be treated as a number. Canceling like that requires thorough justification.
The Infinite Looper (tilper.com) But why can't they cancel each other since they are of the same value? No matter how small they are. Sir can you please explain about this (infinitesimal)? Thanks!
They aren't of the same value. "dg" is just a symbol that represents an infinitely small quantity, just like how ∞ represents an infinitely large quantity. When you see ∞, it doesn't always represent the same value. There is no actual value, in fact, in the same sense that, e.g., 4 has a value. Also, there are different levels of infinity and the symbol ∞ doesn't distinguish between them. ∞ is not a real number, and so you can't do arithmetic operations on ∞ as if it were a real number. Similarly, dg is not a real number, and so you can't do arithmetic operations on dg as if it were a real number. The "cancelation" that occurs as part of dy/dx = dy/dg * dg/dx isn't cancelation in the same sense that 1/2 * 2/3 = 1/3 has cancelation. It's really just convenient notation. For more info about infinitesimals, I suggest googling hyperreal numbers.
I might seem very stupid but I have 1 question:Why at the very beginning of the video the derivative of f(g(x)) is equal to lim t-->x f(g(t))-f(g(x))/t-x) and not lim t-->x f(g(t))-f(g(x))/g(t)-g(x)?Sorry for such absurd question.
+Kenan Mamedov They're actually the same limit. The reason they're the same limit is because g is a continuous function. Therefore, g(t) approaches g(x) as t approaches x. I'm guessing - I think, I don't know, I recorded this almost 5 years ago but this is what I'm thinking now - the reason I chose to write as t approaches x is because x is the independent variable. The one we can control. But g is a function and is dependent upon x. So even though we use g as input to f, the function g itself must first receive an input. And it is this input to g that we need to consider.
+The Infinite Looper (tilper.com) It's just i don't quite get the difference between d[f(g(x))]/dx and f'(g(x)) which is f(g(t))-f(g(x))/g(t)-g(x),I understand that f'(g(x)) times g'(x) is equal to d[f(g(x))]/dx but I don't understand the difference between this and f'(g(x))...they seem identical to me.What am I getting wrong?
+Kenan Mamedov not a bother! So there are really three functions here. f, g, and the composite function f composed with g. df(g(x))/dx is the derivative of the composite function. f'(g(x)) is simply the derivative of f evaluated at g(x) instead of being evaluated at just x. For example, take f(x) = x^2 + 1 and g(x) = x^3. Then f'(x) = 2x. And so f'(g(x)) = 2g(x) = 2x^3. But the composite function is f(g(x)) = [g(x)]^2 + 1 = x^6 + 1, so the derivative of the composite function is 6x^5. (Note that this equals f'(g(x)) * g'(x).)
It doesn't come from anywhere, at least not in the same sense that the product rule or quotient rule (for example) come from somewhere. The piecewise function here is just something we define to help us with the process. Things we define like that don't really need to come from anywhere. No relation to the error function.
Is the proof equivalent if instead of introducing the piecewise function you just do it in two cases where g(x) ≠ g(t) and g(x) = g(t). For the first case show "the limit"= f'(g(x))g'(x). For the second case show "the limit"= 0 but also f'(g(x)) = 0 and therefore f'(g(x))g'(x) = 0*g'(x) = 0, thus "the limit" = f'(g(x))g'(x) when g(x) = g(t). And to then finally conclude: Since "the limit"= f'(g(x))g'(x) when g(x) ≠ g(t) and also when g(x) = g(t), d/dx[f(g(x))] = f'(g(x))g'(x). I think that would make the proof easier to follow for some (introducing a piecewise function at the start of the proof only to really explain it's purpose towards the end may cause some confusion).
Limits in general can't be evaluated in pieces like that. When you take a limit as t approaches x, t takes on infinitely many values and we don't know when g(t) = g(x) and when g(t) ≠ g(x). It won't make sense to say (for example), "Consider lim as t approaches x of [ f(g(t)) - f(g(x)) ] / [ t - x ] but only when g(t) = g(x)" because we don't know anything about g(t) and when it equals g(x). I've mentioned similar statements to this next one a few times in other videos - since we're taking a limit as t goes to x, we only care about what happens when t is "really close" to x. So we can ignore all the other values that are "farther away" from x. But there are still infinitely many values that t takes on and the problem is that, since g is completely arbitrary, we can never guarantee that t can be "close enough" to x so that we would always have g(t) = g(x) or g(t) ≠ g(x).
answerOfstupids If g(t) = g(x), then f(g(t)) = f(g(x)). In other words, if you put the same thing into a function two different times, then you get the same value back both times. Since f(g(t)) = f(g(x)), then f(g(t)) - f(g(x)) = 0.
h goes to 0, we only care what happens when h is "very close" to 0. If we can make h "close enough" to 0 so that g is strictly increasing or decreasing on a tiny interval around x [the interval would be (x-h,x+h)], then yes, we can use the "flawed" proof. The problem is that f and g are arbitrary, so we can't guarantee g has that property. If g doesn't have that property then we can't use the flawed proof. I'm having trouble thinking of a (nontrivial) function without that property right now.
Logged in to say thank you very much! I'm in intro to Real Analysis and was struggling to understand this proof in my textbook (I also noticed there were a couple typos which was really throwing me off). You did a great job of explaining this proof, and I was able to completely rewrite it from memory after having you explain it! I'm in school right now to be a high school math teacher, and I hope to explain mathematics as well as you in a couple years. Thank you!!!
Thanks a lot! Glad to hear this helped. Don't you just love the textbooks with the typos? Always so useful. Best of luck to you!
G'(0) = 1 is correct.
Yep, it's typically these unusual functions that we wouldn't normally think of right away that show us these strange properties, or that show us we can't assume things that seem intuitive, etc.
Changing my previous answer..
If g(x+h)=g(x) for only finitely many values of h near 0, then we can eventually get h "close enough" to 0 so that g(x+h) doesn't equal g(x). In other words, we can find a small enough open interval centered around x such that g(a) and g(b) are different for every a and b in the interval. This means g must be strictly increasing or decreasing on that interval. I think the intermediate value theorem can be used to formally prove that.
Since we take the limit as
It's good to ask these questions. The derivative of f(g(x)) still exists regardless of what F is. But F has to be chosen in such a way that we can use it to actually show the derivative exists. If we choose F(y) to be constant when y = g(x), then this F will not work. That doesn't actually mean the derivative of f(g(x)) doesn't exist, it just means that the chosen F will not help us show the derivative exists.
Retaining that equality you mentioned earlier is important, but we need more
Good question. Short answer: If you define F(y) to be a constant when y = g(x), then F(y) is not guaranteed to be continuous. But F(y) has to be continuous for this proof to work.
More details: In the video between near 15:40 and near 19:40, I show that the limit as t approaches x on the first piece of F(g(t)) is equal to the limit as t approaches x on the second piece of F(g(t)). This equality shows that F(g(t)) is continuous, and F(g(t)) must be continuous or else the conclusion stated
On the surface it feels like there's too much of a dependency between x, r, and z. For the proof to be valid we need x and z to be arbitrary, but z depends on x and r since x < z < r. Making r arbitrary would also be a problem since there's a dependency between x and r.
But I can't come up with any examples where the proof fails. It might require going beyond polynomials.
Very good explaination, absolutely perfect and complete proof according to IGNOU Standard.
Why is it that f(g(t))-f(g(x))/g(t)-g(x)=g(t)-g(x)/t-x? That part of the proof (around 18:45) felt kinda like you were just saying they looked similar and therefore they're identical. However, if f(x) is 3x and g(x)=2x, for instance, then 6t-6x/2t-2x is not equal to 2t-2x/t-x, is it? One simplfies to 6(t-x)/2(t-x)=3, and the other simplifies to 2(t-x)/t-x=2. So, from this definition, I've derived that 3=2, which is a contradiction. This statement would only hold true if f(x) were equal to g(x), which only proves the chain rule for f(f(x)).
Where is my mistake?
Caleb Spiess What did I say/do around 18:45 that makes you think I suggested those two expressions are equal? I watched the video around that time and didn't see any indication of that.
Oh, after watching it like the fifth and sixth times over I realized what you meant lol. f(g(t))-f(g(x))/g(t)-g(x) is literally the limit definition of f'(g(x)). I don't know why I got so confused at that part, but now I understand the whole proof.
Thanks, by the way! The proof on khanacademy is "start with dy/dx and multiply by du/du to get dy/dx=dy/du * du/dx" which was completely unconvincing to me, so this video helped me get a feel for why the chain rule REALLY works.
Is the plan to use the CMVT to say that [f(r) - f(x)]/[g(r) - g(x)] = f'(z)/g'(z), and then apply the hypothesis? If that's the case then we would always need x < z < r, and I'm not entirely sure we can always have that if we also want abs( [f(r) - f(x)]/[g(r) - g(x)] - f(x)/g(x) ) to be less than e/2. For that inequality, we would need x and r to be "close enough" to each other. It may be the case that getting x and r "close enough" might not be possible if we also must have x < z < r.
Do you mean at a specific value of x, or for any value of x?
For a specific x, not necessarily. Consider a variation of the function I mentioned in another response: G(x) = x^3 * sin(1/x) + x if x is not 0, and G(0) = 0. This function is continuous and differentiable at x=0, and it's true that G(0+h)-G(0)=0 for infinitely many values of h near 0, but G'(0) is not 0.
For any x in general, I think the conclusion would still be not necessarily for similar reasons.
I think such a function would have to be oscillating, like G(x) = x^3 * sin(1/x) when x is not 0, and G(0) = 0. This function is continuous and differentiable at x=0.
Also, this function oscillates near x=0, and it oscillates so much that no matter how close h gets to 0, there is always "another" h closer to x=0 so that G(0+h) = G(0).
than that. We need an F that both retains that equality and is continuous. The F chosen in the video meets these requirements.
If we choose F to be constant when y = g(x), then F retains the equality but is not continuous. Since F is not continuous, then the last step in the proof becomes invalid. And to reiterate the earlier point, that doesn't mean the derivative of f(g(x)) doesn't exist, it just means that the F we chose won't help us find the derivative.
Thanks for this nice educative lecture.
Though my question is about how you have defined F(y). You have given condition for g(t) and g(x) but you have not given us the conditions of t and x in this piecewise function, that is for the upper and lower expressions.
Can you please clarify on this.
Thanks in advance
At 11:56 you say that t-x≠0 because t->x. But how will g(t)=g(x) if t-x≠0? I think for the case g(t)=g(x) when t=x. Am i correct?
But even if we could conclude that g'(x) = 0, dividing by g(x+h) - g(x) would still be a problem because 0 * (number) = 0 only if (number) is some finite quantity.
If g(x+h) - g(x) is 0 for infinitely many h near 0, then {f[g(x+h)-f[g(x)]}/{g(x+h)-g(x)} is undefined for infinitely many h near 0, and undefined * 0 is still undefined. It'd basically be 0/0, which is one of the indeterminate forms.
I'm not sure if that's where you were going with that but I wanted to mention it to be thorough.
You say that we don't know what g does, but you seem to assume that g(t) -> g(x) when t -> x. Is this a property of differentiable functions or am I missing something?
Never mind, I realized this is apparent when you consider the epsilon-delta definition of limits.
Gliffie
Sounds like you got it now, but in case anyone else comes here later wondering the same thing, the fact that g(t) -> g(x) when t -> x follows directly from the fact that g is a continuous function (and can be proven using epsilon-delta techniques), and g being continuous follows directly from the fact that g is differentiable.
Took me a while but i finally got there i think! thanks! I have one question though, at 8:36 you say that the g(t) - g(x) cancel out but at the beginning at 1:54 you say they cant cancel out because of infinitesimals. surely infinitesimals would apply here too then? I wld ask you about the product rule at 14:!4 but ur already answering that on another video ahahahah thanks dude
At 1:54 I said the dg can't cancel out. dg isn't the same thing as g(t) - g(x).
More details: Showing something like this doesn't exist in general is much harder than showing that it does exist. In this case, to show that the derivative of f(g(x)) exists, we only need to pick one function F that works.
If we pick one function F that doesn't work, we can't conclude anything yet. If we wanted to go this route to show the derivative of f(g(x)) doesn't exist, we would have to show that the proof wouldn't work for EVERY function F we could choose.
h goes to 0, we only care what happens when h is "very close" to 0. If we can make h "close enough" to 0 so that g is strictly increasing or decreasing on a tiny interval around x [the interval would be (x-h,x+h)], then yes, we can use the "flawed" proof.
The problem is that f and g are arbitrary, so we can't guarantee g has that property. If g doesn't have that property then we can't use the flawed proof.
But I'm having trouble thinking of a (nontrivial) function without that property.
Why do f and g go to 0 as x goes to infinity? That doesn't hold for any nontrivial polynomial or for many types of transcendental functions. If we begin with that as an assumption then it definitely looks like this all works out, but it's a pretty restrictive hypothesis to have.
Hi! I would like to add that f(g(x)), its derivative, is not df/dx. And it's not df/dx= df/dg * dg/dx. But instead it should have been y=f(g(x)) and dy/dx=dy/dg*dg/dx which is equivalent to [f o g]'(x)= f'(g(x))*g'(x). And the dg'es do cancel each other since they are actually just a number (an extremely small one) with limit -> 0 (infinitesimal). Is that correct sir? thanks!
Euvlid W your first comment is right, my mistake.
Your second comment is not correct. Infinitesimals are not real numbers (like 1, 3.7, -12, pi, sqrt(5), etc. are) and cannot be treated as numbers, just like how infinity is not a number and cannot be treated as a number. Canceling like that requires thorough justification.
The Infinite Looper (tilper.com) But why can't they cancel each other since they are of the same value? No matter how small they are. Sir can you please explain about this (infinitesimal)? Thanks!
They aren't of the same value. "dg" is just a symbol that represents an infinitely small quantity, just like how ∞ represents an infinitely large quantity. When you see ∞, it doesn't always represent the same value. There is no actual value, in fact, in the same sense that, e.g., 4 has a value. Also, there are different levels of infinity and the symbol ∞ doesn't distinguish between them. ∞ is not a real number, and so you can't do arithmetic operations on ∞ as if it were a real number. Similarly, dg is not a real number, and so you can't do arithmetic operations on dg as if it were a real number. The "cancelation" that occurs as part of dy/dx = dy/dg * dg/dx isn't cancelation in the same sense that 1/2 * 2/3 = 1/3 has cancelation. It's really just convenient notation. For more info about infinitesimals, I suggest googling hyperreal numbers.
Dude you are the bomb dot com!
I might seem very stupid but I have 1 question:Why at the very beginning of the video the derivative of f(g(x)) is equal to lim t-->x f(g(t))-f(g(x))/t-x) and not lim t-->x f(g(t))-f(g(x))/g(t)-g(x)?Sorry for such absurd question.
Wait...is not it the general formula of this ENTIRE function,right?While f(g(t))-f(g(x))/g(t)-g(x) is the derivative of the outer function,right?
+Kenan Mamedov They're actually the same limit. The reason they're the same limit is because g is a continuous function. Therefore, g(t) approaches g(x) as t approaches x. I'm guessing - I think, I don't know, I recorded this almost 5 years ago but this is what I'm thinking now - the reason I chose to write as t approaches x is because x is the independent variable. The one we can control. But g is a function and is dependent upon x. So even though we use g as input to f, the function g itself must first receive an input. And it is this input to g that we need to consider.
+The Infinite Looper (tilper.com) It's just i don't quite get the difference between d[f(g(x))]/dx and f'(g(x)) which is f(g(t))-f(g(x))/g(t)-g(x),I understand that f'(g(x)) times g'(x) is equal to d[f(g(x))]/dx but I don't understand the difference between this and f'(g(x))...they seem identical to me.What am I getting wrong?
+The Infinite Looper (tilper.com) And thanks for the quick answer,I hope that you won't miss my second question,mate.I'm sorry for bothering you.
+Kenan Mamedov not a bother! So there are really three functions here. f, g, and the composite function f composed with g. df(g(x))/dx is the derivative of the composite function. f'(g(x)) is simply the derivative of f evaluated at g(x) instead of being evaluated at just x. For example, take f(x) = x^2 + 1 and g(x) = x^3. Then f'(x) = 2x. And so f'(g(x)) = 2g(x) = 2x^3. But the composite function is f(g(x)) = [g(x)]^2 + 1 = x^6 + 1, so the derivative of the composite function is 6x^5. (Note that this equals f'(g(x)) * g'(x).)
Thanks, glad to help!
where does the piecewise function come from. Is it similar to the error functions used in other videos?
It doesn't come from anywhere, at least not in the same sense that the product rule or quotient rule (for example) come from somewhere. The piecewise function here is just something we define to help us with the process. Things we define like that don't really need to come from anywhere. No relation to the error function.
between 20:50 and 21:00 is not true, and the step we do right after that is invalid.
this is awesome. thks for helping me understand this rule!
Is the proof equivalent if instead of introducing the piecewise function you just do it in two cases where g(x) ≠ g(t) and g(x) = g(t). For the first case show "the limit"= f'(g(x))g'(x). For the second case show "the limit"= 0 but also f'(g(x)) = 0 and therefore f'(g(x))g'(x) = 0*g'(x) = 0, thus "the limit" = f'(g(x))g'(x) when g(x) = g(t).
And to then finally conclude: Since "the limit"= f'(g(x))g'(x) when g(x) ≠ g(t) and also when g(x) = g(t), d/dx[f(g(x))] = f'(g(x))g'(x).
I think that would make the proof easier to follow for some (introducing a piecewise function at the start of the proof only to really explain it's purpose towards the end may cause some confusion).
Limits in general can't be evaluated in pieces like that. When you take a limit as t approaches x, t takes on infinitely many values and we don't know when g(t) = g(x) and when g(t) ≠ g(x).
It won't make sense to say (for example),
"Consider lim as t approaches x of [ f(g(t)) - f(g(x)) ] / [ t - x ] but only when g(t) = g(x)" because we don't know anything about g(t) and when it equals g(x).
I've mentioned similar statements to this next one a few times in other videos - since we're taking a limit as t goes to x, we only care about what happens when t is "really close" to x. So we can ignore all the other values that are "farther away" from x. But there are still infinitely many values that t takes on and the problem is that, since g is completely arbitrary, we can never guarantee that t can be "close enough" to x so that we would always have g(t) = g(x) or g(t) ≠ g(x).
i dont get it. You can prove anything is equal to anything by just making it equal to zero. 2*0=0. & 500 - 500= 0 so a =2 b=500. ?
I'm not sure which part of the video you thought I used logic like that.
The Infinite Looper in last part where you said its 0 so its same as f(g(x) - f(g(x)).....anyways what is even happening...
answerOfstupids
If g(t) = g(x), then f(g(t)) = f(g(x)). In other words, if you put the same thing into a function two different times, then you get the same value back both times.
Since f(g(t)) = f(g(x)), then f(g(t)) - f(g(x)) = 0.
very well explained! Thanks so much
Thnx sir beautifully explained....
Sure, go ahead.
h goes to 0, we only care what happens when h is "very close" to 0. If we can make h "close enough" to 0 so that g is strictly increasing or decreasing on a tiny interval around x [the interval would be (x-h,x+h)], then yes, we can use the "flawed" proof.
The problem is that f and g are arbitrary, so we can't guarantee g has that property. If g doesn't have that property then we can't use the flawed proof.
I'm having trouble thinking of a (nontrivial) function without that property right now.