Is my proof for the Chain Rule invalid?

[Werg22]

How about you check post 32.

 

[prasannapakkiam]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

 

[Werg22]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

This was the argument but it no longer is. The argument now is that the equality of the ratio and the product of the other ratios as well as the equality the limit of the first ratio with the product of the other ratios are doubtful. But as I said, for the case in which there exist a neighborhood in which the functions are monotonic, there's nothing to be doubted.

 

[JonF]

How about you check post 32.

This is not the definition of the limit of a function, and your explanation was far from formal, and didn’t really what exactly it means to take to limits keeping “equal change”. Did you go over the delta epsilon definition of a limit in your class?

 

[Office_Shredder]

That is it.

Arguments seem to be revolving around whether dy/dx or dy/du can be split up into a fraction. How about we take a look at: f'(x)=lim h--> 0 (f(x+h)-f(x))/h. How did we derive it? We used small values of y and x i.e. dy and dx. Remember that all the equal signs in my proof are the "aprox." signs. The limit in the end ties it all up to a differential. But the limit thing is neccessiarily neccessary...

Yes, someone said'hey, if we use this as a definition, we get the slope for a bunch of nice looking curves. So let's extend this, and define it to be the slope for all curves'

Why? Because for all curves, that includes curves that aren't nice. So yes, you can use this intuitive sense of derivative, and if your curve behaves welll, expect to be right. But the point of the definition is that it's the only way to 100% accurately describe the derivative of a function, no matter which function you're using.

Technically, werg has yet to prove that he's even covered all the cases (and as we can see by his initially missing the case where it hits zero infinite times in any interval, it's not obvious that all the cases are covered).

However, by definition using the definition of a derivative hits all the cases

 

[Werg22]

Notice how these [insert you know what here] keep on attacking my proof without actually putting anything forth. Office_Shredder, you are so laughable. As soon as Mathwonk stepped in, you changed your tune. It's pretty obvious now that you really don't know what you are talking about.