Say a function f is differentiable. Prove whether its derivative is continuous over its domain or not.  On an intuitive level, I believe it should be always continuous over the domain (open intervals) and am thinking of a rigorous proof.
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If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?  My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable.
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Let $f:\mathbb{R}\to\mathbb{R}$ be a map sending closed intervals to closed intervals. Prove that $f$ is continuous or find a counter example.  WLOG we just have to prove continuity at $0$ and we can also assume that $f(0)=0$ and $f(1)=1$ (by considering $x\mapsto (f(x)-f(0))/(f(1)-f(0))$ ).  I build up this question while thinking of the question .
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I need to  Let $f : (a, b) → R$ be differentiable on $(a, b)$. Prove that $f$ is continuous on $(a, b)$ by showing that:  $∀x_0 ∈ (a, b)$ $∀ε &gt; 0$ $∃δ &gt; 0$ s.t.  $(x ∈ (a, b), |x − x_0| &lt; δ) ⇒ |f(x) − f(x_0)| &lt; ε$  My understanding so far:   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.  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  My attempted 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  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  My issue:  My issue may be a non issue, however for the 6 marks this question should be worth my proof seems shallow. Could someone please poke some holes in my proof, or add to it?   Thank you
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I am new to multivariable calculus and my textbook doesn't give out solutions so I'm just wondering how you go about proving something like this? I know that a function is differential at a point $a$ if it's continous at $a$ and and the partial derivatives of $f$ exist near $a$ but I have never actually seen an example. Heres the question:    Assume $f(0, 0) = 0$, and determine whether $f(x, y) = \frac{xy^3}{x^2 + y^4}$ is differentiable at $(0, 0)$.
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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.
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Theorem: If $f$ is finitely differentiable at $c$, then $f$ is also continuous at $c$.  But according to the definition, for a function to be differentiable at $c$, it need not even be defined at $c$, just that it should have a finite value in the vicinity of $c$.    My question: If function is not defined at $c$, how can it be continuous at $c$?  For example, say we have a straight horizontal line function which is broken at $x=1$, the function is defined at all the points except $x=1$, it has a finite value at all the points except $x=1$, since it is having a value in the nearby region of $x=1$ we say that it is differentiable at $x=1$ but can we say that it is continuous at $x=1$?
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Is a Lipschitz function differentiable?  I have been wondering whether or not this property applies to all functions.      I do not need a formal proof, just the concept behind it.    Let $f: [a,b] \to [c,d]$ be a continuous function (What is more - it is uniformly continuous!) And let's assusme that it's also Lipschitz continuous on this interval.    Does this set of assumptions imply that $f$ is differentiable  on $(a,b)$?
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Suppose for real numbers $a&lt;b$ one has a function with continuous derivative $f:[a,b] \to \mathbb{R}$ such that $f(a) = 0$ and there exists a real number $C$ with $$|f'(x)| \leq C|f(x)|$$ for all $x \in [a,b]$. Show that $f(x) = 0$ for all $x \in [a,b]$.  Well, since $f(a) = 0$, we have that $|f'(a)| \leq C|0|$, so $|f'(a)| = 0$. Since $f$ has a continuous derivative, we also know that $f$ is continuous. Since $f$ is continuous on a compact interval, $f$ obtains a maximum, say at $\xi \in [a,b]$. So, $|f'(x)| \leq C|f(\xi)|$. Since the derivative is bounded, we obtain that $f$ is Lipschitz, so $f$ is also uniformly continuous. Suppose to the contrary that $f(\xi) &gt; 0$? The above is pretty much everything I could figure out about $f$, so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course. I think maybe I should try to show that $|f'(x)| = 0$ for all $x$, but I don't know how.
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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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Assume $f$ is a function over $\mathbb{R}$ satisfying $f(x+y) = f(x)+f(y)$ for all $x,y \in \mathbb{R}$. Show that if $f$ is continuous, then $f(x) = cx$ for all $x \in \mathbb{R}.$   I find it hard to use the definition of continuity to derive anything from the equation. I think we could write the definition of the limit in terms to get $$\forall \epsilon,\exists \delta \quad 0&lt;|x-a|&lt;\delta \quad \implies \quad |f(x)-L| &lt; \epsilon.$$ I am not sure where to go from here.
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I have the following question, more general that   Let $U\subset\mathbb{R}^n$ be a domain. If $f:U\longrightarrow\mathbb{R}^m$  ($m&lt; n$) is a continuous function, how to prove that $f$ is not injective.  Any hint would be appreciated.
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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?
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$f(x)$ is continuous and differentiable function defined in $x \geq 0$. If $f(0)=1$ and $f'(x) &gt;3 f(x)$, then prove that $$f(x) \geq  e^{3x}, \forall x \geq 0$$  Could someone give me little hint as how to proceed in this question? I am not able to initiate.
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Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.
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exercise: Suppose that $f:(0,∞) → R$  satisfies $f(x) - f(y) = f(x/y)$  for all $x,y$ in $(0, ∞)$ and $f(1) = 0$.   Prove f is continuous on $(0, ∞)$ if and only if f is continuous at $1$.    (proof→:) Suppose that $f:(0,∞) → R$ satisfies $f(x) - f(y) = f(x/y)$ for all $x,y$ in $(0, ∞)$ and $f(1) = 0$. Suppose f is continuous on (0,∞), then by definition f is said to be continuous at a point $a$ if for every $ε &gt; 0$  there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$.   Then suppose f is continuous at $1$, then by our defined function  f(x) - f(y) = f(x/y) , we get $|x-1| &lt; δ$ implies $|f(x) - f(1)| = |f(x) - 0| = |f(x)| =  &lt; ε$.   (proof ←) :Suppose that $f$ is continuous at $1$, then we need to show $f$ is continuous on every point $a$ on $(0, ∞)$. Thus, by definition, f is said to be continuous at a point $a$ if for every $ε &gt; 0$ there is a $δ &gt; 0$ such that $|x-a| &lt; δ$ implies $|f(x) - f(a)| &lt; ε$. Thus, $|x-a| &lt; δ → |f(x) - f(a)| = |f(x/a)| &lt; ε$. Then $f$ is continuous on on $(0, ∞)$.  Please can anyone please help me verify this makes sense. If not, can someone please give any feedback/hints. Thank you in advance.
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If $f\colon \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$   and $f'(0)$ exists, then $f$ is differentiable on $\mathbb R$.   I know how to show it is continuous but no clue how to show differentiable.
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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.