The question is the nth derivative of $x^{x}$  I have try solving this nth term differentiation of x raised to the power of x. Using $y=x^{x}$ and taking he log of both sides...but I don't know how to differentiate the nth term. Pls help
120834
A few days ago, I was wondering about a series expansion for $$f(x) = \  x^x$$ so I tried taking a couple derivatives. From these I was able to guess at some formulas for the general coefficients of the first few terms of the nth derivative as well as the last term (where I put the terms of the derivative in the order  $$f^{(n)}(x) = c_1x^x + c_2x^{x-1} + c_3x^{x-2}\ + \ ... \ +\ c_nx^{x-(n-1)}$$  Please let me know if there is a general formula for the coefficients as a function of n. In my own fooling around, I found that $$\ln(x)+1$$ appears quite a bit, so I called this $a$ in my work. Since I would like to center my expansion at $1$, and this expression evaluated to $1$, at $x=0$, I was interested in writing my formulas as functions of $n$ and $a$. I found that you get something of the form (this could very well be completely incorrect):  $$f^{(n)}(x) = ax^x + a^{n-2}{\binom n2}x^{x-1} + a^{n-4}(3{\binom n4}-a{\binom n3})x^{x-2}\\ + a^{n-6}(15{\binom n6}-10{\binom n5}a+2{\binom n4}a^2)x^{x-3}\ + \ ... \ +\ (-1)^n(n-2)! \ x^{x-(n-1)}$$  An interesting property that seemed to recur for the various coefficients was  that once I found a recursion formula for the coefficient in question and plugged in my initial value, it always seemed to be the same as taking the integral of the previous coefficient with respect to $a$. That is to say, if I plugged in the nth coefficient (a polynomial of $a$) into my recursion formula for the $(n+1)$th coefficient, this gave me the same result as integrating. In general, as reflected by the few coefficient formulas above,  the formula for the coefficient of $$x^{x-c}$$ where c is an integer such that $$0\le c&lt;n$$ "Stabilized" to a polynomial of a with c terms after the first few applications of the recursion formula. This stabilization occurred when $$n=2c$$ Also, as a direct result of the tentative and incomplete formula I have written above, the first nonzero-value for the coefficient of $$x^{x-c}$$ is $$(-1)^{c+1}(c-1)!$$ and appears in $$f^{(c+1)}(x)$$ If these various properties are indeed true, can anyone explain why this would be the case? I am aware that this function can be expanded using the expansion for $$e^x$$ with $$x\ln(x)$$ in the exponent. This does not strike me as particularly enlightening (let me know if I'm missing something). I am most interested in the coefficients on the binomial coefficients, since these are the only things that stand between me and a general formula for the coefficients (since the form is quite standard). These are: $$1$$ then $$3,-1$$ then $$15, -10,2$$The sum of these coefficients should also give the nth derivative evaluated at zero since $$f(1) = 1$$ and $$a(1) = 1$$  Anyway, please let me know what you think! I am currently stuck due to an inability to brute force my way any further and a complete lack of cleverness. My knowledge of mathematics is quite elementary, so please explain any higher level concepts if they are necessary to explain this problem.
66152
Suppose I knew nothing about the function $e^x$. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to $1$. Then, the next term should be the antiderivative of the first term, giving me $x$. Doing this again would give me $\frac{x^2}2$. Repeating this process over and over again, I would get $$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$ Graphing a few terms of this, I might notice that this looks more and more like an exponential the more terms I graph. If I prove that this function satisfies the exponential relationship $f(x+y) = f(x)f(y)$, I would be able to prove that this series is an exponential function. How would I prove this? After this, how would I prove that the base of this exponential function is $e$, which is defined as $\displaystyle{\lim_{n \to \infty}} (1+\frac1n)^n$?  Edit: After expanding $(1+x)(1+y)$ and $(1+x+\frac{x^2}2)(1+y+\frac{y^2}2)$, I can see how extra terms get taken care of when the next degree is added. However, my second question still stands.
97835
$x, a$ in $\mathbb R^n$, $A$ in $\mathbb  R^{n\times n}$. Compute $d(x^T a)/dx$ and $d(x^T A x)/dx$.  I'm not sure about how to think about these and how to do these. Can someone explain how to derive the expressions for the two?  Finally, what happens when we have $A$ and $X$, BOTH in $\mathbb R^{n\times n}$, and we want to find $dTrace(XA)/dX$?
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.
43106
I do as follows  $(1+x)^n= \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$   differentiating in both sides  $n(1+x)^{n-1}= \binom{n}{1}+2\binom{n}{2}x+3\binom{n}{3}x^2...+n\binom{n}{n}x^{n-1}$  then, for $x=1$  $n \cdot2^{n-1} = \sum_{k=1}^{n}k\binom{n}{k}$  From here, I don't know how to proceed, I have tried to multiply both sides for $(1+x)$ to make $n \cdot2^{n}$ but then the summatory changes, I don't know how to prove the equality.
58222
I have seen many times that the gradient of $x^TA\,x$ with respect to $x$ is $2A\,x$. But how do you find its gradient with respect to $A$?
18097
Taking the factorial function $x!$  I wonder if there is a method to find the first derivative of this function without making any use of the Gamma function (or related integral representations of the factorial). Maybe something like that  $$ \begin{align} \frac{\text{d}}{\text{d} x}\ x! &amp; = \frac{\text{d}}{\text{d} x}\ (x\cdot (x-1)\cdot (x-2)\cdot \cdots) \\\\ &amp; = ((x-1)\cdot (x-2)\cdots) + x\cdot ((x-3)\cdot (x-4)\cdots) + x\cdot ((x-2)\cdot (x-4)\cdots) + \cdots  \\\\ &amp; = (x-1)! + x\cdot N(x) \end{align} $$  What I'm missing is a suitable $N(x)$ to express the remaining terms.  Does such a function exist?  I thought about this, observing term by term:  $$N(x) = \left(\prod_{k = 3}^{n} (x-k)\right) + \left((x-2)\prod_{k = 4}^n (x-k)\right) + \left((x-2)\cdot(x-3)\prod_{k = 5}^n (x-k)\right) + \cdots$$  But still it seems quite messy, isn't it?
90387
Use the definition of the derivative to find $f'(x)$ for $f(x)=\sqrt{x-2}$. I don't even know where to start with this. I have done $\sqrt{x-2}= (x-2)^.5$.  Is this correct?
64823
As many of you probably know, when you take the derivative of $x^x$, you cannot treat it as either a exponential function ($a^x$) or like a function of the form $x^a$.   If you treat it like $x^a$ then you get an aswer of $xx^{x-1}$, which is just $x^x$.  Treat it like $a^x$, and you get $x^xln(x)$.  The correct derivative is $x^x(ln(x)+1)$, which you get by doing implicit differentiation after taking the natural log of both side.  This is the same as $x^xln(x)+x^x$, which happens to be the combination of the "wrong" derivatives you get when treating $x^x$ as the 2 different types of functions I specified earlier.  Is there any real logic behind this or is it coincidence. Also, could this also be applied to other function hybrids like $x^x$?
22772
Given $f = f(u(t), t)$, is it possible to find the following derivative? ($f$,$u \in \mathbb{R}^n$, $t \in \mathbb{R}$)  $$ \frac{\partial f}{\partial||u||^2_2} $$  I am aware of the following derivatives:  $$ \frac{\partial||x||^2_2}{\partial x} = \frac{\partial ||x^Tx||_2}{\partial x} = 2x $$  So I can invert it to give me (elementwise division in the RHS)  $$ \frac{\partial x}{\partial||x||^2_2} = \frac{1}{2x} $$  This step does feel like abuse of notation I find it hard to convince myself of its validity.  If it's right though, I could just apply the chain rule to find  $$ \frac{\partial f}{\partial u} \frac{\partial u}{\partial||u||^2_2} $$  However, knowing that both $f$ and $u$ are a function of $t$, I am somehow doubting my own reasoning. Or does that just mean that the derivative will still be a function of $t$ and I have nothing to worry about? Can anyone shed some light?  UPDATE: I'm now doubting the existence of  $$ \frac{\partial u}{\partial ||u||^2_2} $$  because $u = u(||u||_2^2)$ is not injective. Is the inversion step I did above legitimate? Is there another possible approach?
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!
101867
I can't get any of these on my own and I am attempting to do $x^x$ I had it explained to me for about an hour and I still can't do it on my own. I thought I was supposed to make it into $x(\ln x)$ which should be equivalent to the original term. Then the differentiation should be easy for most people from here.  $$1(\ln x)+x(\frac1x) $$ or $$\ln x+1$$  This is of course wrong but I do not know why.
109313
In the book calculus made easy, page 22 the case of the negative power for   $y=x^{-2}$    $$\begin{align}  y+dy &amp; =(x+dx)^{-2}\tag{1}\\ \\ &amp; = x^{-2}\left(1+\frac{dx}{x}\right)^{-2}\tag{2} \end{align}$$  How do they get from $(1)$ to $(2)$?  And why isn't the binomial function applied to the second line but applied on the third line (the line that comes after line $(2))$?
18482
I've been trying to think for the past few days how one could differentiate $a^x$ based on the definition that $a^n$ is repeated multiplication, $a^{n/m}=(\sqrt[m]a)^n$, and $a^x$ is the completion of the above function by continuity.  With a bit of algebra, the problem quickly reduces to finding the derivative at $0$:  $$\lim_{h\to0}\frac{a^h-1}{h}\tag{1}$$  And that limit's really got me stumped. Since you'll obviously have to use the definition in some way, I thought I'd replace $h$ with $\frac{1}{n}$ and use the $n$-th root definition:  $$\lim_{n\to\infty}n(\sqrt[n]a-1)\tag{2}$$  Of course, just because $(2)$ exists doesn't automatically imply $(1)$ exists, but it might be a first step. Even $(2)$ has me stumped, though.
73841
So taking the derivative of the latter half we get  $$\left(\frac{2}{x^{3}}\right)e^{\frac{-1}{x^{2}}}$$ by chain rule. Im not sure how to find the derivative at x=0, and show that it equals 0 as it only produces more '1/0' expressions. Additionally I assume to show it is not analytic at $x_0=0$, we use the taylor series for $e^{x}$ which is $$1+x+\frac{x^2}{2}+\cdots+\frac{x^{n}}{n!}$$ Where we substitute $\frac{-1}{x^{2}}$ for x to get  $$e^{\frac{-1}{x^{2}}}=1-\frac{1}{x^{2}}+\frac{1}{x^{4}2!}-\frac{1}{x^{6}3!}\cdots$$ So when we plug in x=0 this is invalid and thus is not analytic.  Is my latter attempt sufficient or do I need to add more? And how do I reach the conclusion that the function is differentiable at x=0.
33937
Let $f: \mathbb R \to \mathbb R$ be an infinitely differentiable function such that $f(0)=0$ , $f(x)&gt;0 , \forall x \ne 0$ and   $f^{(n)}(0)=0$ ( the $n$-th derivative ) $,  \forall n \in \mathbb N$ , then is the function $g(x):=\sqrt{f(x)}$ infinitely   differentiable ? I am having difficulty checking whether any $n$-th derivative exist at $0$ . Please help . Thanks in advance
90643
How can we find the derivative of  $y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}$?  so far this is what I have done, and am confused about what to do after?    sorry for the messy handwriting