Can somebody help me solve this equation?  $x\times 10^{\frac{1}{x}}+\frac1x \times 10^x=20$  What I did: I think the left side is decreasing I tried to show that the first derivative  $10^{\frac1x} - \frac{1}{x^2} 10^x - \frac{1}{x}(10^{\frac1x} - 10^x) \log(10)$ is negative, but I don't know how. How to do it?
120892
Solution of equation $\displaystyle \frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$   $\bf{My\; Try::}$ Clearly Here $x&gt;0$, Now Using $\bf{A.M\geq G.M}$  So Here $\displaystyle x\cdot 2014^{\frac{1}{x}}&gt;0$ and $\displaystyle \frac{1}{x}\cdot 2014^x&gt;0$ .  So $\displaystyle \left(\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2}\right)\geq \left(x\cdot 2014^{\frac{1}{x}}\cdot \frac{1}{x}\cdot 2014^x\right)^{\frac{1}{2}} = \sqrt{2014^{\left(x+\frac{1}{x}\right)}} = 2014$   And equality Hold , When $\displaystyle x = \frac{1}{x}\Rightarrow x= 1$  Can we solve it  without Using $\bf{A.M\geq G.M}$  If Yes, Then please explain me.  Thanks
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 $(2))$?
35880
I'm struggling with finding derivative of f(x)=arth(lnx).  I've done following:  x=th(y)  f'(x)=(arth(lnx))'=1/(thlnx)'=1/(1/(xch^2(lnx)))=(x(ch^2(lnx)))/(ch^2x-sh^2x)=(ch^2(x(lnx)))/(ch^2(1-th^2x))=(x(lnx))/(1-x^2)  so in conclusion I get f'(arth(lnx))=(x(lnx))/(1-x^2)  but I think it's not correct. Any ideas?
67871
I have this HW where I have to calculate the $74$th derivative of $f(x)=\ln(1+x)\arctan(x)$ at $x=0$. And it made me think, maybe I can say (about $\arctan(x)$ at $x=0$) that there is no limit for the second derivative, therefore, there are no derivatives of degree grater then $2$. Am I right?
99220
Find $f$ if $$ f(x)+f\left(\frac{1}{1-x}\right)=x $$ I think, that I have to find x that $f(x) = f\left(\frac{1}{1-x}\right)$ I've tried to put x which make $x = \frac{1}{1 - x}$, but this equation has no roots in real numbers.
94328
Using first principles, differentiate $$f(x)=(x)^{\frac{3}{4}} $$  So$$\lim_{h\to 0}\frac{f(x+h) -f(x)}{h} = \lim_{h\to 0}\frac{(x+h)^{\frac{3}{4}} -(x)^{\frac{3}{4}}}{h} $$  I cant seem to solve this question. I'm supposed to get the   output $\frac{3}{4}x^{-\frac{1}{4}} $
100646
This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this?
70134
I want to calculate the derivative of $|x|^\alpha$ with respect to $x$, where $1&lt;\alpha&lt;2$.  $\frac{d|x|^\alpha}{dx}=\alpha|x|^{\alpha-1}\mathrm{sign}(x)$  Is it correct?  And is it true that  $\lim_{x \to 0} \alpha|x|^\alpha\mathrm{sign}(x)=0$?  What with the following derivative $|x|^\alpha\rm{sign}(x)$. I would be grateful for any hints.
25962
I have got this problem on limits  $$ \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2}$$ What I am doing here is that taking log and then applying L-Hospital's Rule $$y= \lim_{x \rightarrow 0}\left(\frac{\sin(x)}{x}\right)^\frac{1}{x^2} $$ then $$\ln(y)=\lim_{x \rightarrow 0} \frac{1}{x^2}\ln\left(\frac{\sin(x)}{x}\right)$$ now it becomes $\frac{0}{0}$ type , but problem is how to take derivative here? when differentiating $\ln\left(\frac{\sin(x)}{x}\right)$ it will become $\frac{x}{\sin(x)}$ then we'll have to differentiate one more time like we do in case of $f(f(x))$ ? $\sin(x)$ and $x$ will be separately differentiated or together ?
47323
Using $$\frac{f(x+h)-f(x)}{h}$$ how would I find the derivative of $f(x)= \sqrt{x}+1$   The answer is $$\frac{1}{2\sqrt{x}}$$ but I keep getting $$\frac{1}{2 + 2\sqrt{x}}$$
13008
Can someone help me solve the equation $x^2 - 1 = e^x$ ?  I tried taking the natural logarithm of both sides but I don't know where to go from there..  I got:  $\ln(x^2 -1) = x$ But I don't know how to solve it from here. Any help please?
115659
I'm trying to solve a limit problem but I've never encountered the one like this before.  $$ \lim_{t\to 0}\left(\frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)$$  I multiply the right side by $\frac{t\sqrt{1+t}}{t\sqrt{1+t}}$ and combine the terms to get:  $$ \lim_{t\to 0}\left(\frac{1 - t\sqrt{1+t}}{t\sqrt{1+t}}\right)$$  I can factor out the $t\sqrt{1+t}$ and I am left with:  $$1 - 1 = 0$$  This seems wrong to me. I can't explain why, maybe I just feel that getting a zero after all that work seems lame. Am I following the right steps?
121303
Solving $\frac{1}{x} &lt; 4$ gives me $x &gt; \frac{1}{4}$. The book however states the answer is: $x &lt; 0$ or $x &gt; \frac{1}{4}$.  My questions are:  Why does this inequality has two answers (preferably the intuition behind it)?  When using Wolfram Alpha it gives me two answers, but when using $1 &lt; 4x$ it only gives me one answer. Aren't the two forms equivalent?
117830
How to integrate $\frac{1}{x\log x}$?  Could you give me some ideas on how to integrate this? thanks.  i've tried setting $u=(\log x)^{-1}$.   $\dfrac{\mathrm du}{\mathrm dx} = x^{-1}$  But it didnt work...
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?
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
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 $f(x)&lt;0$.