For the function $F(x) = \int_0^x |\cos(1/u)| du$, I want to determine if the derivative exists at $x=0$ and find $F'(0)$.  I figured this out when the absolute value is removed using integration by parts:  $$\frac{F(h) - F(0)}{h} = \frac{1}{h}\int_0^h\cos(1/u)du = \frac{1}{h}\int_{1/h}^\infty\frac{\cos(x)}{x^2}dx \\= - h \sin(1/h) + \frac{1}{h}\int_{1/h}^\infty\frac{2\sin(x)}{x^3}dx = -h \sin(h) + \frac{1}{h}O(h^2)$$  So $F'(0) = 0$ here.  This breaks down for $F(x) = \int_0^x \cos(1/u)du$.  Thank you for any help.
122021
Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance
51766
If $f'(c)$ exists then $\lim_{x\to c} f(x)=f(c)$   My attempt is below. Did I do wrong? Since I don't really understand it. I thought $f'(x) = f'(c)$ so $f'(x)-f'(c)=0$. Am I on the right track? Many thanks!  \begin{align} f(x)-f(c)&amp;=\frac{f(x)-f(c)}{x-c}\cdot(x-c) \\ \lim_{x\to c}[f(x)-f(c)]&amp;=\lim_{x\to c}\left[\frac{f(x)-f(c)}{x-c}\cdot(x-c)\right] \\ \lim_{x\to c}[f(x)-f(c)]&amp;=0 \\ \lim_{x\to c} f(x)&amp;=f(c) \end{align}
37731
I'm not entirely sure what I want to ask here, so please bear with me!  I think the explanation they give us in school for how finding the antiderivative by substitution works is: $$\int f(g(t))g&#39;(t)dt=(Fg)(t)=F(g(t))=F(u)=\int f(u)du$$  But I never really understood the equality $F(u)=\int f(u)du$. Why can we behave as if the identity $u=g(t)$ doesn't exist, compute the integral $\int f(u)du$, and then 'substitute' $u$ in the result yielding the correct answer? In a sense, it feels like the variable $u$ switches from being 'meaningful' in $F(u)$ (where it stands for $g(t)$) and being insignificant in $\int f(u)du$, since as I perceive it the "variable name" here is meaningless and we could say $ \int f(u)du = \int f(k)dk = \int f(x)dx = ... $ all the same. Another way to ask the same thing: why $(Fg)(t)=\int f(u)du$ and not, say, $(Fg)(t)=\int f(x)dx$? Where am I getting confused?  Maybe someone can explain this better than my high school teacher? Is there a 'formal' explanation for this? Thank you a lot!
29673
I was messing around with the definition of the derivative, trying to work out the formulas for the common functions using limits. I hit a roadblock, however, while trying to find the derivative of $e^x$. The process went something like this:  $$\begin{align} (e^x)' &amp;= \lim_{h \to 0} \frac{e^{x+h}-e^x}{h} \\ &amp;= \lim_{h \to 0} \frac{e^xe^h-e^x}{h} \\ &amp;= \lim_{h \to 0} e^x\frac{e^{h}-1}{h} \\ &amp;= e^x \lim_{h \to 0}\frac{e^h-1}{h} \end{align} $$  I can show that $\lim_{h\to 0} \frac{e^h-1}{h} = 1$ using L'Hôpital's, but it kind of defeats the purpose of working out the derivative, so I want to prove it in some other way. I've been trying, but I can't work anything out. Could someone give a hint?
81051
I am trying to take the derivative with respect to $a$ of some function $I(a)=\int_{0}^{\infty}f(a,x)dx$.  I would like to make sure that I am using the  correctly.  Various web sources indicate a set of conditions that must hold for $f(x,a)$ and $\frac{\partial f(x,a)}{\partial a}$ when integration is done over infinite region.  From reading  (see Theorem 10.3 on page 13) the conditions that $f(x,a)$ and $\frac{\partial f(x,a)}{\partial a}$ must obey are:   $f(x,a)$ and $\frac{\partial f(x,a)}{\partial a}$ are continuous over $x\in[0,\infty)$ and around $a$ that we are interested in. There exists an integrable function (over $x$) $g(x)$ such that $|\frac{\partial f(x,a)}{\partial a}|\leq g(x)$. There exists an integrable function (over $x$) $h(x)$ such that $|f(x,a)|\leq h(x)$.   Integrable here means $\int_{-\infty}^{\infty}g(x)dx&lt;\infty$.  However,  seems to omit condition 3 above.  I am wondering which source is correct.  If there are "both correct", when is condition 3 necessary?
115607
Define the real valued function $f(x)$ by $f(x) = e^{-1/x^2}$ if $x \ne 0$ and $0$ if $x = 0$.    How do you show that $e^{-1/x^2}$ is differentiable at $0$? How do you show that this function is infinitely differentiable?
3599
Assume this function has a value $0$ at $x=0$.  If the right hand derivative of a derivable function is equal to the right hand limit of the derivative function (same for the left), why aren't functions like the one above continuously differentiable?  For a function to be continuous, limits should be  finite and equal to value at that particular point.  For this function derivative using first principle yields $0$.  The RHD, LHD is $0$, I think. So why isn't the derivative continuous?  For finding the RHD and LHD, I have seen answers resorting to using known derivatives. My question is: why don't we use the first principle itself? Why do we get different answers then?
40778
In integrals, we often make 'u' substitutions, which involve writing $u=f(x)$, and then also writing $du=f'(x)dx$. How valid is this? I've heard that you aren't allowed to do things like 'break' up derivatives like they do to solve first order differential equations. So can you do this with integrals, or is it also a mnemonic?
82206
When deriving $x^x$, why can't you choose $u$ to be $x$, and find $\dfrac{d(x^u)}{du} \dfrac{du}{dx} = x^x$? Or you could go the other way and find $\dfrac{d(u^x)}{du}\dfrac{du}{dx}$, giving $\ln(x)\cdot{x^x}$? Both methods seem to be equally wrong.
40411
So far I've done this, but I don't know if it will help.  Let $\int_0^xf(t)dt=F(x)$.  Now, $1=\int xf(x)=F(x)x|_0^1-\int_0^1F(x)dx=1-\int_0^1 F(x)dx$. Then, $\int_0^1 F(x)dx=0$.  But I don't know how to keep going.
28896
Given the function: $$f(x) = \begin{cases} x\cdot\sin(\frac{1}{x}) &amp; \text{if $x\ne0$} \\ 0 &amp; \text{if $x=0$} \end{cases}$$  Question 1: Is $f(x)$ continuous at $x=0$?  Question 2: What is the derivative of $f(x)$ at $x=0$ and how do I calculate it?
2565
How can I find $f(0)$, $f'(0)$, and $f'(x)$ given that  $f(x+y)=f(x)+f(y)+x^2y+xy^2$  and  $\lim_{x\to0}\frac{f(x)}{x}=1$.
81925
Question: Let $f:[0,1]\to\mathbb{R}$ be a continuous function such that $f(0)=0$ and $$\int_0^1f(x)dx=0.$$ Prove that $\exists a, b\in(0,1)$ such that $$\int_0^{a} xf(x)dx=0\text{ and }\int_0^bxf(x)dx=\frac{b^2f(b)}{2}.$$  My approach: Let $g:[0,1]\to\mathbb{R}$ be such that $$g(x)=x\int_0^xf(t)dt-\int_0^xtf(t)dt, \forall x\in[0,1].$$ By the first fundamental theorem of calculus we can conclude that $g$ is differentiable on $[0,1]$ and $$g'(x)=\int_0^xf(t)dt, \forall x\in[0,1].$$ Also, observe that $g(0)=0$ and $g(1)=-\int_0^1tf(t)dt$. Thus, by applying MVT to the function $g$ on the interval $[0,1]$, we can conclude that, $\exists c\in(0,1)$ such that $$g'(c)=\int_0^cf(t)dt=-\int_0^1tf(t)dt.$$ Observe that clearly three cases are possible, i.e, either$$\int_0^cf(t)dt&lt;0\text{ or }\int_0^cf(t)dt=0\text{ or }\int_0^cf(t)dt&gt;0.$$ Now let $h:[0,1]\to\mathbb{R}$ be such that $$h(x)=\int_0^xtf(t)dt, \forall x\in[0,1].$$ Please note that the part highlighted below is wrong, but still I have included it to just demonstrate my thinking process, as it might be of some help to others trying out this problem.  Observe that if $\int_0^cf(t)dt&lt;0$, then, $h(1)&gt;0$. This also implies that $\exists$ an open interval $(d,e)\in[0,c]$, such that $f(t)&lt;0, \forall t\in(d,e)$. Now select any point $c_1\in(d,e)$. Applying MVT to the function $h$ on the interval $[0,c_1]$, we can conclude that $\exists c_2\in(0,c_1)$ such that $$h'(c_2).c_1=f(c_2).c_2.c_1=h(c_1)-h(0)=h(c_1)&lt;0.$$ Now $h(c_1)&lt;0$ and $h(1)&gt;0$. Thus, by IVT we can conclude that $\exists a\in(c_1,1)\subseteq(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ A similar reasoning for the case when $\int_0^cf(t)dt&gt;0,$ shows that $\exists a\in(0,1)$, such that $$h(a)=\int_0^af(t)dt=0.$$ Now finally if $\int_0^cf(t)dt=0$, then we will have $h(1)=0$. Now if $f$ is identically equal to $0$ on $[0,c]$, then clearly $tf(t)=0, \forall t\in[0,c]\implies h(x)=0, \forall x\in[0,c].$ Thus choosing any point $x\in(0,c]$ and setting it as $a$, we will have $h(a)=0$ and we will be done in that case. Now if $f$ acquires both positive and negative values on $[0,c]$, then we can conclude that $\exists c_1,c_2\in(0,c)$, such that $f(c_1)&gt;0$ and $f(c_2)&lt;0$. Also, let us assume WLOG that $c_2&gt;c_1$.  I have not been able to make any significant approach other than this. Can someone help me out with this problem? Please note that a solution using integration by parts might not be possible, since $f$ is not a differentiable function.
103505
I was wondering if there are functions for which $$f'(x) &gt; f(x)$$ for all $x$. Only examples I could think of were $e^x - c$ and simply $- c$ in which $c &gt; 0$. Also, is there any significance in a function that is always less than its derivative?    Edit: Thank you very much for all the replies. It seems almost all functions that apply are exponential by nature...  Are there more examples like - 1/x?  Again are there any applications/physical manifestations of these functions? [for example an object with a velocity that is always greater than its position/acceleration  is always greater than its velocity]
103853
The question asks us to find the function $f(x)$ with the given information     Let $f:\mathbb R   \rightarrow  \mathbb R$ such that $f'(0)=1$ and $f(x+y)=f(x) + f(y) + (e^{x+y})(x+y)-xe^x-ye^y+2xy$ where $x,y ∈\mathbb R$. Then determine $f(x)$.   Also,It will be great if someone could explain the general approach used in such questions.  Thanks in advance.
75824
I have this function $$f(x) = x\cos(1/x)$$ defined for all $x \geq 1$.      How do I prove that    $$f(x+2)-f(x)&gt;2$$    for all $x\geq 1$?   I found $$f'(x)= \cos(1/x) + \frac{\sin(1/x)}{x}$$ and $$f''(x)=-\frac{\cos(1/x)}{x^3}.$$ How do i use this information?
94633
I am trying to find the maximums or minimums of $$f(x)=|x-1|+|x-2|+|x-3|$$ (if there exist).  My attempt: First I compute the derivative and tried to find critical point, i.e, $f'(x) = \frac{x-1}{|x-1|} + \frac{x-2}{|x-2|} + \frac{x-3}{|x-3|}=0$, first I noted that this derivative doesn't exist in 1,2,3 for the absolute value. And the are no critic point (Is that correct?)  Then I stuck here because I don't know if there is a maximum, I think that it is not exist but how can I justified and I believe that there is a global minimum, but this occurs only if I found a critical point, some help pls.