Here's a problem on differentiability at a point and continuity of the derivative function at the same point. I'm stating the problem and presenting my solution to it. I'm not entirely confident on my solution (especially about interchanging the limits). So I'd greatly appreciate if someone checks the solution and tell me if there's any gap in my arguments. Thank you.     The Problem : Let $f:(a,b) \to \mathbb{R}$ is continuous on $(a,b)$ and is differentiable except possibly at $c \in (a,b)$. Assume that $\lim_{x \to c} f'(x)$ exists. Show that $f'(c)$ exists and $f'$ is continuous at $c$.   My Approach : Given that $\lim_{x \to c}f'(x)$ exists, we can write \begin{align} \lim_{x \to c}f'(x)&amp;=\lim_{x \to c} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ &amp;=\lim_{h \to 0} \lim_{x \to c} \frac{f(x+h)-f(x)}{h} ~~\text{[Since the double-limit exists]}\\ &amp;=\lim_{h \to 0} \Big[\frac{1}{h} \cdot \lim_{x \to c}\big\{f(x+h)-f(x)\big\}\Big]\\ &amp;=\lim_{h \to 0} \frac{f(c+h)-f(c)}{h} ~~\text{[Since $f$ is continuous]}\\ \end{align} Since $\lim_{x \to c}f'(x)$ exists, so does $\lim_{x \to c} \frac{f(c+h)-f(c)}{h},$ i.e. $f'(c)$. Clearly, $\lim_{x \to c}f'(x)=f'(c)$. Hence $f'$ is continuous at $c$. $~\blacksquare$  P.S. The problem is from an exercise consisting of Mean Value Theorem (MVT)-related problems. So it is highly likely that MVT should play a role here, which makes me suspect this is a comparatively simple, probably too-simple-to-be-correct solution.
122767
Suppose that $\mathcal f$ is continuous on a closed interval and $\mathcal v$ is a point in its interior. $\mathcal f$ is differentiable when $\mathcal x\not=v$, and suppose that $\,\displaystyle{\lim_{x\rightarrow v}f'(x)}\,$ exists.  Show that $\mathcal f$ is differentiable at $\mathcal v$.  Attempt at a solution  I want to show that   $$ \lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$   exists. Based on the fact that   $$ \lim_{x\rightarrow v}f'(x)=\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}$$  exists. Using the fact that $\mathcal f$ is continuous, I think if somehow I could interchange the limits $$\mathcal {\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}}={\lim_{h\rightarrow 0}\ \lim_{x\rightarrow v}\frac {f(x+h)-f(x)}{h}}=\lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$ then the result comes out. But I don't think that I could do that, based on continuity. Other than that, I am not sure how to approach this.
123870
Let $f : \Bbb R \to \Bbb R$ be the function defined by $f(x)= \frac {\sin x}{|x|+ \cos x}$. Then A.$f$ is differentiable at all $x \in \Bbb R$. B.$f$ is not differentiable at $x =0$. C.$f$ is differentiable at $x=0$ but $f'$ is not continuous at $x=0$. D.$f$ is not differentiable at $x=\frac {\pi}{2}$.  Using the standard definition of derivative, I get that $f$ is differentiable at $0$ as well as $\frac {\pi}2$. This is as follows, $\lim_{h \to 0} \frac {f(0+h)-f(0)}{h}=1=\lim_{h \to 0} \frac {f(0-h)-f(0)}{-h}$ and $\lim_{h\to 0} \frac {f(\frac {\pi}2 +h)-f(\frac {\pi}{2})}{h}=0=\lim_{h \to 0} \frac {f(\frac {\pi}2 -h)-f(\frac {\pi}2)}{-h}$ using L'Hospital rule at one stage. Hence I eliminate options B and D. Since $0$ was the only doubtful point to check about differentiability, I choose option A as the answer. I regret a little about not knowing how to check validity of option C. Can you tell me how to prove it wrong?
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
91704
This is an exercise from Remmert's Theory of Complex functions.  Let $D\subset \mathbb{C}$ be a domain and $f:D\rightarrow \mathbb{C}$ a real-differentiable function. Assume that the following limit exists:  $ \mathrm{lim}_{h\rightarrow 0} \left|  \frac{f(c+h) - f(c)}{h}   \right|.$  Show that either $f$ or $\overline{f}$ is complex-differentiable.  I've tried showing that $\frac{\partial f}{\partial \overline{z}} = 0$ or $\frac{\partial \overline{f}}{\partial z} = 0$ by using the fact that there exist continuous functions $g$ and $h$ such that in $D$ one can write   $f(z) = f(c) + (z-c)g(z) + (\overline{z} - \overline{c})h(z)$   and that  $g(c)= f_{z}(c)$ and $h(c) = f_{\overline{z}}$ and then plugging this into the limit above. Does this approach works and I just can´t see how to do it? Can someone give a hint or a guideline solution to this?
54805
Let $F:\mathbb{R^2}\rightarrow \mathbb{R}$ be a function, such as : $$F(x,y)=\begin{cases}        0 &amp; (x,y)=(0,0) \\       \frac{xy}{x^2+y^2} &amp; (x,y)\neq (0,0)    \end{cases}$$      Is $F$ differentiable at $(0,0)$?   My attempt :  Consider the limit, $\lim_{(x,y)\rightarrow (0,0)} F(x,y)$ along the line $y=x$. The limit equals to $\frac{1}{2}$. However, $F(0,0)=0$. Hence $F$ is not continuous at $(0,0)$ which implies that $F$ isn't differentiable as well.  Is this approach okay ?
11127
I've been trying to solve the following problem:  Suppose that $f$ and $f&#39;$ are continuous functions on $\mathbb{R}$, and that $\displaystyle\lim_{x\to\infty}f(x)$ and $\displaystyle\lim_{x\to\infty}f&#39;(x)$ exist. Show that $\displaystyle\lim_{x\to\infty}f&#39;(x) = 0$.  I'm not entirely sure what to do. Since there's not a lot of information given, I guess there isn't very much one can do. I tried using the definition of the derivative and showing that it went to $0$ as $x$ went to $\infty$ but that didn't really work out. Now I'm thinking I should assume $\displaystyle\lim_{x\to\infty}f&#39;(x) = L \neq 0$ and try to get a contradiction, but I'm not sure where the contradiction would come from.   Could somebody point me in the right direction (e.g. a certain theorem or property I have to use?) Thanks
86812
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.
29487
Problem : Given $f(x) = x + |x|$ for what values of $x$ is $f$ differentiable?   For the sake of generality, let's assume that it is unknown to us that $|x|$ is not differentiable at $x = 0$  Attempted Solution :  Using the definition of differentiability, a function is differentiable over an interval $I$ $\text{iff}$  $$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}, \ \ \forall x\in I$$  Now implicitly this definition of differentiability requires $$\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} = \lim_{h\to 0^-} \frac{f(x+h)-f(x)}{h}$$  Therefore $f$ will be differentiable only when  $$\lim_{h\to 0^+} \frac{(x + h + |x+h|)-(x + |x|)}{h} = \lim_{h\to 0^-} \frac{(x - h + |x-h|)-(x + |x|)}{h}$$  But it is unclear what to do next as $x$ is just arbitrary $$ \begin{equation} \begin{aligned} |x+h| &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq -h\\ -x-h &amp; \text{if} &amp; x &lt; -h \end{cases} \\ &amp;= \begin{cases}x+h &amp; \text{if} &amp; x \geq 0\\ -x-h &amp; \text{if} &amp; x &lt; 0 \end{cases} \end{aligned} \end{equation}$$    A Wrong Solution :  We could take the derivative of $f$, (and evaluate the domain of the derivative $f'(x)$ to values for which it is defined), using the rules of differentiation, but we would get a wrong answer.  $$f'(x) = \frac{|x| + x}{|x|}$$ $$\implies f'(x) = \begin{cases} 2 &amp; \text{if} &amp; x \geq 0 \\ 0 &amp; \text{if} &amp; x &lt; 0 \end{cases}$$  This implies the derivative $f'(0)$ exists, when it does not as per the definition of differentiability. Why is that so?    Questions:   How can a solution be found using the definition of differentiablity? Why does taking the derivative of $f$, $f'(x)$ and evaluating it's domain, not give the correct values of $x$ for which $f$ is differentiable?
16048
mathematicians! I want to ask to all wise people about a problem I met at the quiz to obtain some ideas. The problem is following. It may not be accurate since the problem is dependent on my memory   Let $f : [0, \infty) \rightarrow \mathbb{R}$ and $f$ is differentiable. $|f'(x) |\; \leq\  |f(x)|$ for all $x \in [0, \infty)$. $f(0) = 0$. Then prove that $f(x) = 0$ for all $x \in [0,\infty)$.  At the quiz, I applied the Mean Value Theorem and the Cauchy-Schwarz inequality.   For any $x$, $|f(x)| = |\{f(x) - f(0)\}(x-0)| = |f'(c_{1})(x)|$ for some $c_{1} \in (0,x)$, and then  $$|f'(c_{1})(x)| \leq |f'(c_{1})||x| \leq |f(c_{1})||x|$$ by Cauchy-Schwarz and given assumption. By doing so consecutively, we can obtain the inequality that  $$ |f(x)| \leq |f(c_{n})||x||c_{1}|\ldots |c_{n-1}| $$ for any $x \in [0,\infty)$. As $\{c_{i}\}, \; i\in \mathbb{N}$ is decreasing sequence and $f$ is continuous, $|f(c_{n})|$ can be arbitrary close to $0$ when $c_{n} \rightarrow 0$. Thus the inequality above can be bounded by $\epsilon$.  I think my answer is right, but a bit uncertain. Could you give me a certainty? Thank you very much.
2111
I've been asked to solve the following problem and I believe that I have gone about it correctly but I would appreciate a second look from someone with more experience.  Question: If $f+g$ and $f$ are differentiable at $a$, must $g$ be differentiable at $a$?  Since $f+g$ and $f$ are differentiable at $a$, $f+g$ and $f$ are continuous at $a$.  Observe   \begin{align*}(f+g)^{\prime}(a)-f^{\prime}(a)&amp;=\lim_{x\to a}\frac{(f(x)+g(x))-(f(a)+g(x))}{x-a}-\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\\ &amp;=\lim_{x\to a}\frac{(f(x)+g(x))-(f(a)+g(x))-(f(x)+f(a))}{x-a}\\ &amp;=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}\\ &amp;=g^{\prime}(a). \end{align*}  Thus, if $f+g$ and $f$ are differentiable at $a$, then $g$ must be differentiable at $a$  I feel that I went about this proof correctly, however the move from two limits to a single limit seems like it might have been done incorrectly.
27937
is $$f(x) =\begin{cases} x^2\sin(\frac{1}{x}) \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$ a continuous function specially at point x=0?  And why being derivable its derivative is not continuous at x=0 since  $$f`(x) =\begin{cases} 2x\sin\frac{1}{x}-\cos\frac{1}{x} \mbox{ for } x\neq 0 \\ 0  \mbox{ for } x= 0\end{cases}$$?  I suspect this as to do with being derivable a function f $\mathbb{R}^n\rightarrow\mathbb{R}^m$ that don't imply its partial derivatives are continuous .this I don't get it so far.however partials derivatives if continuous is derivable.  Can you provide me a more general case than just one variable where f is derivable but its partials are not continuous?
110056
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?
79431
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' &gt; 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).  Is $f$ increasing on $(a,b)$? Must $f'(c)$ be zero, or can it be negative?  Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points. And I think $f'(c)$ must be zero: If $f'(c) &lt; 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) &lt; 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) &gt; 0$, a contradiction.
85008
Let f be a function continuous on [0,1] and differentiable on (0,1)  My attempt:  I have tried solving using Rolle's theorem for f'(c) = 0  I let c&lt;x then by Mean Value Theorem, f'(c) = (f(x)-f(0))/(x-0) = f(x)/x since f'(c) = 0, f(x) = 0 Hence f'(c) = f(x)  Similarly, I let c>x and by Mean Value Theorem again, f'(c) = (f(1)-f(x))/(1-x)       = f(x)/(1-x) since f'(c) = 0, f(x) = 0 Hence f'(c) = f(x)  I feel like this proof is wrong because I don't know how to continue with the case where c=x
10735
I am stuck with the following problem that says :     let $f \colon [a,b] \to \mathbb{R}$ be continuous on $[a,b]$    and $f''$ exist for all $x \in (a,b)$. Let $a \lt c \lt b$.    Prove that there exists a point $\xi \in (a,b)$    such that $$f(c)=\frac{b-c}{b-a}f(a)+\left( \frac{c-a}{b - a} \right)f(b)+\frac12(c-a)(c-b)f''(\xi).$$   Can someone explain in details? Thanks in advance for your time.
32572
I did not know what to search to see if this existed elsewhere. But, I could not find it. Here's the question:  Does there exist a continuously differentiable function, $f: [0,1] \rightarrow [0, \infty)$, such that the following hold? (i) $f(0) = 0$ (ii) $f'(x) \le c f(x)$ for all $x$ ($c$ is a fixed constant) (iii) $f \not\equiv 0$  This was a fun problem someone asked me a long time ago. I have always been convinced there is no such function, but I cannot prove it. I have mainly tried using the mean value theorem (in multiple forms), and re-wording the problem in terms of the integral of a continuous function to no avail. I spent some time trying to use (ii), the definition of derivative, and squeeze theorem, but no luck. After much time trying to prove non-existence, I spent a little time trying to find such a function... also with no luck. This problems ability to avoid being solved has since taken away the original fun and replaced it with a twinge of annoyance.
61316
The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.