I noticed something funny. If you differentiate $x^x$ treating the exponent as a constant, you get $xx^{x-1}=x^x$. If you treat the base as a constant, you get $x^x \ln{x}$. If you add these two bizzare and incorrect derivatives of $x^x$, you get $x^x(1+\ln{x})$, which is correct!  Is it merely a weird and funny coincidence or is it a part of a deeper result?  (I'm checking it for $(x^x)^x$, but it will take a while :) )
117450
Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.)
4432
Does $x^{y^z}$ equal $x^{(y^z)}$? If so, why?   Why not simply apply the order of the operation from left to right? Meaning $x^{y^z}$ equals $(x^y)^z$?  I always get confused with this and I don't understand the underlying rule. Any help would be appreciated!
71071
Find the $n^{th}$ derivative of $$f(x) = e^x\cdot x^n$$  If i am not wrong i have following  $1^{st}$ Derivative: $e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x$  $2^\text{nd}$ Derivative: $e^x\cdot n\cdot (n-1)\cdot x^{n-2} + 2 \cdot e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x$  $3^\text{rd}$ Derivative: $e^x\cdot n\cdot (n-1)\cdot (n-2)\cdot x^{n-3} + 3 \cdot e^x\cdot n \cdot (n-1)\cdot x^{n-2} + 4 \cdot e^x \cdot n \cdot x^{n-1} + 2 \cdot x^n\cdot e^x$  From here how do I calculate the $n^{th}$ derivative?  Thanks. :)
105513
Representation similar to $Z'X(X'X)^{-1}X'Z$ frequently appear to e.g. 2SLS.  I think that $Z'X(X'X)^{-1}X'Z = Z'XX^{-1}X'^{-1}X'Z = Z'(XX^{-1})(X'^{-1}X')Z = Z'Z$. So why it seems that in the context of e.g. 2SLS, the representation can not be simplified? What is the mistake I made?
67803
Alright, so the idea of an exponent, $x$, is that you are multiplying its base by itself $x$ number of times. With base $5$ and $x=3$, we have that $5^3$ = $5 \cdot 5 \cdot 5$  I understand that the logarithm with base $a$ of $x = c$,  tells us that  $$a^c = x$$  and for $c =$ positive; values for $x$ are greater than $1$, and for $c =$ negative; values for $x$ are less than $1$,  and for $c = 0$, values for $x$ are...$1$.  So in short, I understand how, by means of observation of the graph of $f(x) = \log x$, we can see that $f(1) = 0$, BUT, I see no other way to understand why $x^0 = 1$, apart from the graph and everything around that very point.  I honestly cannot get my head around the idea, "$5$ times itself $0$ times... is one".  Is it that there is no fundamental answer for this but that we simply know by the graph? Or can I truly understand $x^0 = 1$ on its own?
87756
This actually triggered me in my mind from . After some playing around I notice that the relation $a^{\log_b x} = x^{\log_b a}$ is true for any valid value of $a,b$ and $x$. I am very inquisitive to see how this holds ?
13531
Consider the following:   $1 = 1^2$ $2 + 2 = 2^2$ $3 + 3 + 3 = 3^2$   Therefore,   $\underbrace{x + x + x + \ldots + x}_{x \textrm{ times}}= x^2$   Take the derivative of lhs and rhs and we get:   $\underbrace{1 + 1 + 1 + \ldots + 1}_{x \textrm{ times}} = 2x$   Which simplifies to:   $x = 2x$   and hence    $1 = 2$.     Clearly something is wrong but I am unable pinpoint my mistake.
13763
I can't seem to find a good way to solve this.  I tried using L'Hopitals, but the derivative of $\log(n!)$ is really ugly. I know that the answer is 1, but I do not know why the answer is one.  Any simple way to go about this?
85412
How find the asymptotic behavior for $x(n)$  if $x^{x^x} = n$?  I supposed that $x = O(\log\log{n})$ and took logarithm two times.  So I get $x = O(\frac{\log\log{n}}{\log\log\log{n}})$  Is it right? How to improve this estimate?
41370
How to solve following equation: $x^x = C$ as $x, C \in \mathbb{R}_{&gt;0}$. I tried to use exponent and such that: $x\ln x = \ln C$.
42702
Of course, it is easy to see, that the integral (or the antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for $\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is $x^{\alpha+1}/(\alpha+1)$.  I was wondering if there is an intuitive (probably geometric) explanation why the case $\alpha=-1$ is so different and why the logarithm appears?  Some answers which I thought of but which are not convincing:   Taking the limit $\alpha=-1$ either from above or below lead to diverging functions. Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.
114444
The inconsistency I see between mathematical subjects is really confusing me. I understand that it isn't possible for $e^x$ to be less than zero for real $x$, which is probably why they say that the integral is $\ln(|x|)$.  Before I ramble on too much, I just want to ask: Is there a set of guidelines to follow to help me choose whether to let the integral of $\frac{1}{x}$ equal to $\ln(x)$ or $\ln(|x|)$?  Thanks, Aralox
74620
Okay, maybe I'm missing something obvious but consider $h_n(x) = x^{ 1+ \frac{1}{2n-1}}$ on the set $[-1,1]$. Now consider a fixed $x \in [-1,1]$, we have:  $$\lim_{n\rightarrow \infty} h_n(x) = x \lim_{n \rightarrow \infty} x^\frac{1}{2n-1} = x$$  But my book says it's $|x|$? Why?
107208
$\log x =Cx^4$ has only one root. Find C.  I don't know how to solve this problem. Do you take derivative on both sides?  I am thinking C equals 0. Am I correct on that?
64549
These days, the standard way to present differential calculus is by introducing the Cauchy-Weierstrass definition of the limit. One then defines the derivative as a limit, proves results like the Leibniz and chain rules, and uses this machinery to differentiate some simple functions such as polynomials. The purpose of my question is to see what creative alternatives people can describe to this approach. The nature of the question is that there is not going to be a single best answer. I have several methods that I've collected which I'll put in as answers to my own question.  It's not reasonable to expect answers to include an entire introductory textbook treatment of differentiation, nor would anyone want to read answers that were that lengthy. A sketch is fine. Lack of rigor is fine. Well known notation and terminology can be assumed. It would be nice to develop things to the point where one can differentiate a polynomial, since that would help to illustrate how your method works and demonstrate that it's usable. For this purpose, it suffices to prove that if $n&gt;0$ is an integer, the derivative of $x^n$ equals $0$ at $0$ and equals $n$ at $1$; the result at other nonzero values of $x$ follows by scaling. Doing this for $n=2$ is fine if the generalization to $n&gt;2$ is obvious.
111392
I am having trouble with this function  $$f(x) = e^{2x}(2x-1)+1$$  I found the domain, domain is $\mathbb R$, it is neither even nor odd function.   I found that $f(x)=0  \Leftrightarrow  x=0$, because it is obvious. I am having trouble with finding the sign of this function. That "+1" is making my life miserable. I can easily find the sign of $e^{2x}(2x-1)$, but what do I do with the constant?   Do I find the minimum of $e^{2x}(2x-1)$ and then add constant to it, and then check the sign?   That seems like too much work, considering that this function is actually a numerator of the first derivative of function that appeared in other graphing assignment. There has to be a better way...  The original graphing assignment was $$g(x)=\ln(e^{\frac{2x}{1-e^{2x}}})$$ You can ignore this one when writing the answer. I included it for the sake of context.   EDIT: By finding the "sign" of a function I mean finding the intervals for which $f(x)&gt;0$ and $f(x)&lt;0$.
83384
This is what we are taught: $$5^{-2} = \left({\frac{1}{5}}\right)^{2}$$  but I don't understand why we take the inverse of the base when we have a negative exponent.  Can anyone explain why?