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anonymous
 5 years ago
Let f(x) = x1, then what are the points at which f is not differentiable?
anonymous
 5 years ago
Let f(x) = x1, then what are the points at which f is not differentiable?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you think typing here is not easy, you may go where we did yesterday

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please provide a little details, as I am new to this thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one second, i'm just finishing up elsewhere I will

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0iam, can i get back to you soon on this? i need to go do a few things. i shouldn't be too long.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@amit7808, can you help?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thanks for coming back

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, should I say, what I understand about the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Sorry, had to clear some programs.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, it's okay. You want to find those points where it's not differentiable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have a composition of two functions that lose differentiability at certain points. You have\[g(x)=x\]and\[f(x)=x1\]then\[(f )o (g)(x)=f(g(x))=x1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Firstly, do you know why the absolute value function loses differentiability?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you do, I can move on.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, I don't have any idea

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay. A function is differentiable at a point x on a domain if its limit from all directions exists, which means, among other things, it must have the same value whether you approach from the left or the right (on an interval, which is what you have here).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you look at the absolute value function, x, it's a line of gradient 1 from x=0 to infinity and a line of gradient 1 from x=0 to infinity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have then the following:\[y=x\]for \[x \in [0,\infty)\]and \[y=x \]\[x \in (\infty,0]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I understand that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(i need to eat while i do this)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So do you want to come back later?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it's ok. may be a little slower

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, the slope of the function x as you approach 0 from the right is 1, while the slope of the function x as you approach from the left is 1. That is, the limit of the definition of the derivative is different depending on your direction. We say that the function is not differentiable at that point. Here, that point is zero.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The derivative of x IS defined at any OTHER point because we can ALWAYS find a small enough interval in which to approach x so that the limit is the same from both directions. It's only at zero here that the function is not differentiable.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah. So you can see what I'm saying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so the point where it is differentiable is that where it has the same slope while approaching from both side?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok then its clear. Please proceed

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0At x=0, this falls down.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, to the crux of this, the function g(x) defined above is not differentiable at the point x=0. It means then, that f(g(x)) will not be continuous there also.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.e.\[x1\]won't be differentiable at x=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, that's half of it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just let me eat some more.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok let me speak then, while you eat

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So from what you said, I can see that the other point would be 1 from the function f(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean 1 would be the point where the curve will not be differentiable

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0When you look at x1, you have to now consider where the composite is not differentiable. We have another absolute function, and so it won't be differentiable at the points where\[x1=0\]That is, where\[x=1\]i.e. when \[x= \pm 1\]since either value can satisfy x=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So your points are 0, +1, 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I didn't understand the line "so it won't be differentiable at the points where x−1=0"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, let u=x1. Then \[x1=u\]If I ask you where u is not differentiable, where would you say that would take place?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i.e. where is u not differentiable?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In general, what are things to consider while solving this type of problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You have a function of a function...as in 'chain rule', but it's just composition of functions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean problems dealing with differentiability

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Start from the 'insideout'. I look for functions within functions and start with the one that is nested the most. You then move out, but keep note of what's going on (e.g. here you might need to think of x1 = u, say, so don't slip up, until you get experience).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You also need to look for places where a function ceases to exist, as in the step function. Just look it up and you'll see the function moves up in steps, but the endpoint on each step isn't included and the function jumps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Be familiar with common functions, like the absolute value function.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0But what are the things to consider with problems involving differentiability ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also, nothing beats a plot!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Just what I said: anywhere where the function goes through a sharp change (e.g. absolute) and anywhere where a function ceases to exist or be continuous.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, have you heard of the exam STEP

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually I am preparing for it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyhow I have another question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/updates/4d9abb520b248b0b471ce9e2?source=email#/updates/4d9ac0cf0b248b0bdc1fe9e2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, its not just for cambridge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Many other accepts it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, here is my next question http://openstudy.com/updates/4d9abb520b248b0b471ce9e2?source=email#/updates/4d9ac0cf0b248b0bdc1fe9e2
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