What a great proof of the product rule, very easy and simple. In wikipedia there is a one adding 0s to the definition, which make it more long and tedious
At 1:04 you establish that lim_{x-> x_0} \delta_{f,x_0}(x) needs to exist with a specific value \delta_{f,x_0}(x_0) , but in the first implication the requirement for differentiability is just that lim_{x-> x_0} \delta_{f,x_0}(x) exists. Why these two stamens are equivalent ? I think that the statement at 1:04 requires more (a limit whit a specific value) and the first stament requires just the limit exists. What am I missing ?
Sorry, I get in trouble again :'(. Suppose that the difference quotient function (f(x)- f(x_0) / x- x_0) is almost everywhere 0 but 1 at x=x_0 by the first definition the function is differentiable at x_0 because the limit exists and it's 0, but by the second definition this difference quotient function can't be extended to a continuos function at x_0, then it's not differentiable ?
Great video as always! One question: Let f(x) be a function that is equal to 1 at x=0 and zero elsewhere. According to the first line in the video, wouldn't f(x) be differentiable at x=0 since the limit of the quotient exists?
Remember that the linear aproximation of f(x) is f(x0) + f'(x0)*(x-x0) If you have the product of two function and do a linear aproximation f(x)*g(x) = f(x0)*g(x0) + (f*g)'(x0)*(x-x0)... But also you can multiply both linar aproximation of f and g... (f(x0) + f'(x0)*(x-x0)) * (g'(x0) + g'(x0)*(x-x0))... Given that the two fórmulas are equal, it's just a thing of substrate repeat terms and see that.. (f*g)' = f'*g + g'*f + f'*g' and the last term is omitted because converge to 0
![Real Analysis 39 | Derivatives of Inverse Functions [dark version]](http://i.ytimg.com/vi/0cktpu5Hxjk/mqdefault.jpg)
What a great proof of the product rule, very easy and simple. In wikipedia there is a one adding 0s to the definition, which make it more long and tedious
Thank you very much!
Thanks a lot for another amazing Video! :D
At 1:04 you establish that lim_{x-> x_0} \delta_{f,x_0}(x) needs to exist with a specific value \delta_{f,x_0}(x_0) , but in the first implication the requirement for differentiability is just that lim_{x-> x_0} \delta_{f,x_0}(x) exists. Why these two stamens are equivalent ? I think that the statement at 1:04 requires more (a limit whit a specific value) and the first stament requires just the limit exists. What am I missing ?
Maybe you are missing the continuity? :)
@@brightsideofmaths Yes, I got it now. Thank you so much, your videos are really helpful !
Sorry, I get in trouble again :'(. Suppose that the difference quotient function (f(x)- f(x_0) / x- x_0) is almost everywhere 0 but 1 at x=x_0 by the first definition the function is differentiable at x_0 because the limit exists and it's 0, but by the second definition this difference quotient function can't be extended to a continuos function at x_0, then it's not differentiable ?
@@diegoharo8305 It think you should recall the definition of a limit again. What you describe at the beginning cannot happen.
Great video as always! One question:
Let f(x) be a function that is equal to 1 at x=0 and zero elsewhere. According to the first line in the video, wouldn't f(x) be differentiable at x=0 since the limit of the quotient exists?
Thanks for the question. The limit of the quotient does not exist as a real number. You calculate (0 - 1)/x for x to 0.
Will it be the last video of Real Analysis of this playlist?
No, the series is still in production :)
I am using this series as a supplement for my college course (I liked your teaching style),
Will be waiting for more videos in this playlist : >
Can you explain the last step of the proof the the product rule in a little more detail? Specifically, why does the term with both deltas equal zero?
Not that deltas vanish but the factor in front (x-x0). We just need continuous functions to do the limit process.
Remember that the linear aproximation of f(x) is f(x0) + f'(x0)*(x-x0)
If you have the product of two function and do a linear aproximation f(x)*g(x) = f(x0)*g(x0) + (f*g)'(x0)*(x-x0)...
But also you can multiply both linar aproximation of f and g...
(f(x0) + f'(x0)*(x-x0)) * (g'(x0) + g'(x0)*(x-x0))...
Given that the two fórmulas are equal, it's just a thing of substrate repeat terms and see that..
(f*g)' = f'*g + g'*f + f'*g' and the last term is omitted because converge to 0
So serious version of differentiation is taught AFTER uniform convergence? Never seen any order like this. Is this what German universities do?
This is what I do :) I think this is a good order because I can highlight the importance of uniform convergence much better.
I actually think uniform convergence _should_ be taught before differentiation, at least in a real analysis course.
Uniform convergence of derivative preserves differentiability
Infinity ♾️ infinity ♾️ ∞
Please, change the title. This is part 35!!!
Thanks :)
@@brightsideofmaths You are gratefully welcome.