${\bf definition.}$ A function $f$ defined on open set $A \subseteq  \mathbb{R}$ is said to be ${\bf strong \; differentiable}$ at $a \in  A$ if       $$ \lim_{(x_1,x_2)  \to (a,a) \\  x_1 \neq x_2 } \dfrac{ f(x_1)-f(x_2)  }{x_1-x_2} = f^{*}(a)$$      exists and is finite and we call $f^*(a)$ the strong derivative of $f$   at $x=a$   How is this notion different than the usual derivative?  I mean, if we set $x_2 = a$, we get our usual derivative...  What is the point of such definition? what is motivation for this?
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Definition: Let $f$ be a real valued function. We say $f$ is $\mathbf{strongly}$ $\mathbf{differentiable}$ at $x = a$ if the following limits exists and is finite:  $$ \lim_{x \to a, y \to a, x \neq y} \frac{ f(x)-f(y)}{x-y} = f^*(a) $$  and we can $f^*(a)$ the strong derivative of $f$ at $a$. Why is this definition of derivative different than the usual one? What is the main crucial point to understand here that makes it different?
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Let $f(x)=|x|^r$ for a given real number $r$. Show that $f$ has first order weak derivatives on the unit ball $B_1(0)\subset \mathbb{R}^n$ provided that $r &gt; 1-n$.  Does anyone have an idea on how to prove this?
119987
Let $f:(a,b)\to \mathbb{R}$ be a strictly increasing function. Does the limit $\lim_{x\to a^+}f(x)$ necessarily exist and is a real number or $-\infty$?  If so, is it true that $\ell=\lim_{x\to a}f(x)\le f(x) \ \ \forall x\in (a,b)$? Please provide proofs.
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If $F$ is a function and $U$ is a set, then what does $F|U$ mean?   In , the Inverse Function Theorem, the notation is used  Also, the file in stating the Lemma 2.14 states that $A \in U$, what does this mean? A set is an element of another set? What is $A$?  EDIT: Could someone kindly explain to me what is going on in Lemma 2.16. How does $\delta$ come into play? I dont see the role in the lemma. I also don't see why this proves one-to-one
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For example if my function is $f:\{1\}\longrightarrow \mathbb{R}$ such that $f(1)=1$.  I have the next context:  1) According to the definition given in Spivak's book and also in wikipedia, since $\lim_{x\to1}f$ doesn't exist because $1$ is not an accumulation point, then the function is not continuous at $1$ (Otherwise it should be $\lim_{x\to 1}f=f(1)$).   2) According to  , as far as I can understand a function is continuous at an isolated point.  I don't understand.   Edit:   Spivak's definition of limit: The function $f$ approaches to $l$ near $a$ means $\forall \epsilon &gt; 0 \; \exists \delta &gt; 0 \; \forall x \; [0&lt;|x-a|&lt;\delta\implies |f(x)-l|&lt;\epsilon]$ Spivak's definition of continuity: The function $f$ is continuous at $a$ if $\lim_{x\to a}f(x)=f(a)$
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$f:\mathbb{R} \to  \mathbb{R}$ be defined as  $f(x)$= $x^2 , x \in \mathbb{Q} \\ 0 \ \ , x \in \mathbb{Q^c}\\ $  How to show that $f$ is differentiable at $x=0.$  I want to try this approach.  here  $f'(x)=$ $2x  \in \mathbb{Q}  \\  0  \in \mathbb{Q^c} $  since $lt \\ x \to 0+$$f(x)$ = $lt \\ x \to 0-$$f(x)= 0$   so $f$ is differentiable at $x=0$
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The definition of a derivative is the slope of a function tangent to a point. It is also defined as $$\lim_{h\to0} \frac{f(x+h)-f(x)}{h}$$ If we apply this to $f(x)= |x|$, we get that it is $\lim\limits_{h\to 0} \dfrac{|h|}{h}$, which is undefined. However, if we look at the graph of $|x|$, we see that there can exist a tangent line at x=0, with slope 0. So why is the derivative undefined instead of 0?
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I have a question, to aid my understanding, about the proof that differentiability implies continutity.$\mathstrut$    Differentiability Definition  When we say a function is differentiable at $x_0$, we mean that the limit:  $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  Continuity Definition  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  Theorem: Differentiability implies Continuity: If $f$ is a differentiable function at $x_0$, then it is continuous at $x_0$.  Proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  We have therefore shown that, using the definition of continuous, if the function is differentiable at $x_0$, it must also be continuous.  My Question  The proof seems to execute the following steps:   Assume the function is continuous at $x_0$ Show that, with little algebra, we can change this into an equivalent question about differentiability at $x_0$. With this little bit of algebra, we can show that if a function is differentiable at $x_0$ it is also continuous.   What I am slightly unsure about is the apparent circularity. In my mind it seems to say, if a function is continuous, we can show that if it is also differentiable, then it is continuous. Rather than what I was expecting, namely, if a function is differentiable, we can show it must be continuous.  Hopefully my confusion is clear. Any help will be greatly appreciated.
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The question I'm trying to answer is:  Let $I$ be an interval and let $f:I\rightarrow\mathbb{R}$ be differentiable on $I$. Show that if the derivative $f'$ is never 0 on $I$, then either $f'(x)&gt;0$ for all $x$ or $f'(x)&lt;0$ for all $x$.  I know I can apply Darboux's Theorem to solve the question.  My Attempt: Let $I=[a,b]$. If $k$ is a number s.t. $f'(a)&lt;k&lt;f'(b)$, then there exists a point $c$ in $(a,b)$ s.t. $f'(c)=k$. Because $f'\neq0$, $0&lt;f'(a)&lt;k&lt;f'(b)$.  I'm really stuck.. Could somebody help me out here? Thank you.
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Let $A$ be a nonempty open subset of $\mathbb{R}$.   Consider a function $f : A \rightarrow \mathbb{R}$. Given that $f$ is continuous, what is the probability that it is differentiable? I suspect it is $0$.
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I want to prove the definition of limit of a function from definition (1) to definition (2).  (1) A function $f(x)$ is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for every sequence $\{x_n\}$ with limit $x^*$, the limit $\lim_{n-&gt;\infty}f(x_n)$ exists and is equal to $f^*$. (2) A function is said to have $f^*$ as the limit as $x$ tends $x^*$ iff for each $\epsilon &gt;0$, there exists a $\delta_\epsilon &gt;0$ such that $|f(x)-f^*| &lt;\epsilon \forall |x^*-x|&lt;\delta_\epsilon$  I am confused about the steps. Suppose I am given an $\epsilon &gt; 0$ and $x^*$. I have to show the existence of a $\delta_\epsilon &gt;0$ such that for all $x$ satisfying  $|x^*-x|&lt;\delta_\epsilon$ I can construct a sequence $x_n$ which converges to $x^*$ and $x_n=x$ for some $n \geq N$. Then I can say that $f(x_n)-f^*=f(x)-f^* &lt; \epsilon$. I think I may have made some mistake about the steps.  My main confusion is the following : Suppose I choose $\delta_\epsilon=\epsilon$. Then if a sequence $\{x_n\}$ converges to $x^*$ then $x_n=x$ (where $x$ satisfies $|x-x^*|&lt;\epsilon$) for some $n\geq N$. Now, $\lim_{n-&gt;\infty}f(x_n)$ exists implies there exists an $M$ such that $\forall n\geq M$, $f(x_n)-f^* &lt;\epsilon$. Now, I need to show that for some $n \geq M$ indeed $x_n=x$. How do I show that ? This $M$ and $N$ are different.
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If $f : \mathbb{R} \to \mathbb{R}$, define $C(f) = \{ x : f \text{ is continuous at } x \}$ and $D(f) = \{ x : f \text{ is differentiable at } x \}$. I have seen it proved that:   $C(f)$ is a $G_\delta$ set. For any $G_\delta$ set $A \subset \mathbb{R}$, there exists a function $f : \mathbb{R} \to \mathbb{R}$ such that $C(f)=A$.   I suspect there is a related characterization for $D(f)$. I see that a related question was asked at , where the accepted answer implies that $D(f)$ is a $G_{\delta \sigma \delta}$ set.   Is this optimal, that is, does there exist $f$ such that $D(f)$ is $G_{\delta \sigma \delta}$ and not $G_{\delta \sigma}$? If so, can an example be given? Conversely, given any $G_{\delta \sigma \delta}$ set $A \subset \mathbb{R}$, does there exist $f$ such that $D(f)=A$?
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(a) Let a function $g:I\to\mathbb{R}$ be continuous at $x_0\in I$. Prove that the function $f$ given by $f(x)=(x-x_0)g(x)$ is differentiable at $x_0$.  (b) Let a function $f:I\to\mathbb{R}$ be differentiable at $x_0\in I$ and $f(x_0)=0$. Prove that there exists a function $g:I\to \mathbb{R}$ that is continuous at $x_0 \in I$ such that $f(x)=(x-x_0)g(x)$.    Thought:  (a)  By $Fermat's$, since $g$ is continous, if $g'(x_0)=0$ then $g(x)$ is differentiable?  but how does it connect back with $f(x)$?  (b)  use the definition of every differentiable function is continuous?  Thus $f(x)$ is continuous and also $(x-x_0)$ so $g(x)$ too?  Thank you!!
113348
If $f:\mathbb{R}^+\to\mathbb{R}$ is monotonic and differentiable, does it follow that $f$ is continuously differentiable?  (This question arose from discussion here: )
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I don't really understand why a discontinuous function cannot be differentiable.  In Stewart's Calculus, the definition of a function $f$ being differentiable at $a$ is that $f'(a)$ exists. Earlier it gives the definition of the derivative as $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. It also has a theorem that if $f$ is differentiable at $a$, then $f$ is continuous at $a$. There is also a definition that a functions $f$ is continuous at a number $a$ if $\lim_{x\to a}f(x)=f(a)$.  Take for example the very simple function:  $$ f(x)=\begin{cases} x+1 &amp; x\geq0, \\ x &amp; x&lt;0.  \end{cases} $$  It is discontinuous at $x=0$ (the limit $\lim_{x\to 0}f(x)$ does not exist and so does not equal $f(0)$), but if I find the derivative using the limit above, I get the left and right limits to equal $1$. So therefore, the derivative exists.  According to the book, the function shouldn't be differentiable at $x=0$ as it has a discontinuity (continuity is a necessary condition of differentiability). What am I doing/understanding wrong?
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Suppose $f$ is a continuous, strictly increasing function defined on a closed interval $[a,b]$ such that $f^{-1}$ is the inverse function of $f$. Prove that,   $$\int_{a}^bf(x)dx+\int_{f(a)}^{f(b)}f^{-1}(x)dx=bf(b)-af(a)$$   A high school student or a Calculus first year student will simply, possibly, apply change of variable technique, then integration by parts and he/she will arrive at the answer without giving much thought into the process. A smarter student would probably compare the integrals with areas and conclude that the equality is immediate.  However, I am an undergrad student of Analysis and I would want to solve the problem "carefully". That is, I wouldn't want to forget my definitions, and the conditions of each technique. For example, while applying change of variables technique, I cannot apply it blindly; I must be prudent enough to realize that the criterion to apply it includes continuous differentiability of a function. Simply with $f$ continuous, I cannot apply change of variables technique.  Is there any method to solve this problem rigorously? One may apply the techniques of integration (by parts, change of variables, etc.) only after proper justification.  The reason I am not adding any work of mine is simply that I could not proceed even one line since I am not given $f$ is differentiable. However, this seems to hold for non-differentiable functions also.  I would really want some help. Pictorial proofs and/or area arguments are invalid.
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If a function $f$ is continuous at every $c$ point in $[a, b]$ then we have for every $\epsilon$ we can find a $\delta$ such that whenever $|x-c|&lt;\delta$, we have $|f(x)-f(c)|&lt;\epsilon$.      Now, in this definition, $c$ varies in the interval $[a, b]$, and also $x$ varies in $[a, b]$, then what is the difference between $x$ and $c$?