I can not figure out how to write “find $f(x)$ such that $\frac{df}{dx}$ is equal to the inverse function of $f(x)$”, or $f^{-1}(x)$)  I am not looking for the answer to this problem, just how you would write it in notation with $y^\prime$ and such.
120798
Question 1: Study the existence of  $C^1$ function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\forall x\in\mathbb{R},\mbox{ } f\circ f'(x)=x.$    Question 2: Study the existence of  differentiable function  $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfing $\forall x\in\mathbb{R},\mbox{ } f\circ f'(x)=x.$    Question 3: Study the existence of  $C^1$ function $f : \mathbb{R} \rightarrow \mathbb{R}$ satisfying $\forall x\in\mathbb{R},\mbox{ } f'\circ f(x)=x.$    For question 1, such a function can not exist because : f' must be injective and since f 'is continuous, f' must be strictly monotonous. For example, If we assume that f 'is strictly increasing  We can show that $\displaystyle \lim_{x\rightarrow -\infty}f'(x)=-\infty$ and $\displaystyle \lim_{x\rightarrow +\infty}f'(x)=+\infty$ which implies that f ' is surjective .  with a simple argument it shows that f is injective ( if $f(x)=f(y)$ by surjection of $ f'$ , we have $f'(a)=x $ and $f'(b)=y$ for some real $ a,b $ , thus implie $a=f(f'(a)=f(x)=f(y)=f(f'(b)=b$  so $ x=y$).   the continuity of f proves that f est strictly monotonous. For example, if we suppose f strictly inreasing, we must have $f'&gt;0$. this contradicts the surjectivity of $f '$  For Question 2 , I need help
14433
What are the functions that are their own inverse?  (thus functions where $ f(f(x)) = x $ for a large domain)  I always thought there were only 4:  $f(x) = x , f(x) = -x , f(x) = \frac {1}{x} $ and $ f(x) = \frac {-1}{x} $   Later I heard about a fifth one     $$f(x) = \ln\left(\frac {e^x+1}{e^x-1}\right) $$ (for $x &gt; 0$)  This made me wonder are there more? What are conditions that apply to all these functions to get more, etc.
80981
I am aware there are many questions similar to this on MSE but I am having trouble following any of the solutions given.   I have the function given by   $f(x) = \dfrac{xy}{x^2+y^2}\;$ for $(x,y)\neq (0,0)$ and $f(0,0)=0$  I have calculated the partial derivatives and found:   $$\frac{\partial f}{\partial x} = \frac{y(-x^2+y^2)}{(x^2+y^2)^2}, \quad \frac{\partial f}{\partial y} = \frac{x(-y^2+x^2)}{(y^2+x^2)^2}$$  Now I need to show the partial derivatives exist for $(x,y) = (0,0)$.  I am also asked to show it is not continuous at $(x,y)=(0,0)$, based on other answers I've seen it seems like this follows from solving the first part but I fail to see how that follows through as well.   I could just copy the solutions with my function in place as I have seen a lot of answers using the definition of the derivative, but realistically I do want to understand what the thinking behind this is.   Any help would be appreciated.
10474
What is the inverse of   $$f(x)=\sin(x)+x.$$  I thought about it for a while but I couldn't figure it out and I couldn't find the answer on the internet.  What about  $$f(x)=\sin(a \cdot x)+x$$  where $a$ is a known real constant.  Thank you for taking the time to read this question! Sorry if this has been asked before...
109489
How do we find the inverse of the function $f:(-1,1)\to \Bbb R$ by $f(x)=\frac{x}{1-x^2}$?  The problem has been posted  and the answer below says "To show that $f$ has a continuous inverse, you just need to show that $f$ is an open map..."   Is it impossible to find the inverse analytically?
91199
I asked 3 professors at my university and none gave me a clear cut answer, but instead merely told me qualities of this notation. Here is what I understand so far from what they told me:   1)Treat the top variable as as variable when finding the derivative 2)Treat the bottom variable as a constant when finding derivative 3)It it said "Find x with respect to y", but what exactly does that mean? What does it mean for something to be in respect to something else?  It seems like $\frac{dx}{dy}$ notation changes according to values in the problem. For instance, If $y = x^3 + 2x$ and $\frac{dx}{dt} = 5$, find $\frac{dy}{dt}$ when $x=2$. Why do the values of $\frac{dx}{dy}$ change in this problem and how do I solve this?
56113
Well, yesterday at a Mathematics exam, i had to find $\frac{dy}{dx}$ of a cotangent inverse function $$ y=\text{arccot}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-sin x} }\right]$$ My Approach: $$ \text{Multiply and Divide by}\frac{1}{\sqrt{1+\sin x}}$$ $$ \text{but}\space \frac{\sqrt{1-sinx}}{\sqrt{1+\sin x}}=\sec x-\tan x$$ $$\text{so} \space \space y=\text{arccot}\left[\frac{1+\sec x-\tan x}{1-\sec x+\tan x}\right]$$  $\text{Then no idea, Anwers is 1/2}$  Can anyone Give me Full answer with explanation
112613
If $f(t) = 1+ \frac{1}{2} +\frac{1}{3}+....+\frac{1}{t}$,  Find $x$ and $y$ such that $\sum^n_{r=1} (2r+1)f(r) =xf(r) -y$
55206
Find a function $f(x)$ such that:  $$f(x)f(1/x)=f(x)+f(1/x)$$  with $f(4)=65$.  I have tried to let $f(x)$ be a general polynomial: $$a_0+a_1x+a_2x^2+\ldots a_nx^n$$  which leaves $f(1/x)$ as:  $$a_0+a_1{1\over x}+a_2{1\over x^2}+\ldots + a_n{1\over x^n}$$  On comparing the coefficients of both sides, we see that:  $$2a_0=(a_0)^2+(a_1)^2+(a_2)^2+\ldots+(a_n)^2$$  And  $$a_1=(a_0a_1)+(a_1a_2)+ \ldots +(a_{n-1}a_n)$$  I don't know how to proceed further. I know I need to compare coefficients and come to a conclusion based on their values, but I don't see what to do next.
51991
$f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2}  } , \frac{y}{x^{2}+y^{2}  } \right)  $  How do I find inverse function? I have trouble here, because there are 2 variables and I am not sure how to do it?
40589
Can you help me find the inverse function for $y = x^3 + x$?  This question was posed at the beginning of AP Calculus, so we can't use any math beyond precalc.  Thanks!
83001
I am trying to solve the following functional equation, and could use some help.$$ 2f(x)=f(ax)$$  For some $a\in\mathbb{R}$. By repeated adding $2f(x)$ together we notice that $$2nf(x)=f(a^nx).$$  Also $$2mf(a^{-m}x)=f(x)\Rightarrow \frac{1}{2m}f(x)=f(a^{-m}x).$$  Putting it all together I have for all $m,n\in\mathbb{N}$, $$\frac{n}{m}f(x)=\frac{2n}{2m}f(x)=f(a^{n-m}x)$$  I am unsure of how to proceed from here.
9551
Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$  Now I want to find the value of $F(2)$. I can do it in 2 ways:   Put $x=2$ and solve the function. It will give:  $$F(2)=\frac{0}{0}$$ which is not defined. Solve $F(x)$ first and then put $x=2$.   $$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$  It will give $${F(2)=\frac{0}{4}}$$ which is zero.   How can zero equal not defined?
71394
when I apply $f^{-1}$ to both sides $f(f(x))=f(x)$ $(1)$ we get $f(x)=x$ $(2)$.  are $1$ and $2$ equivalent?
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.
97092
In  thread, the question was to find a $f: \mathbb{R} \to \mathbb{R}$ such that  $$f(f(x)) = f(x) + x$$  (which was revealed in the comments to be solved by $f(x) = \varphi x$ where $\varphi$ is the  $\frac{1+\sqrt{5}}{2}$).    Having read about  shortly before though, I came up with this train of thought:  $$f(f(x)) = f(x) + x$$ $$\Leftrightarrow f^2 = f^1 + f^0$$ $$f^2 - f - f^0 = 0$$  where $f^n$ denotes the $n$'th iterate of $f$.  Now I solved the resulting quadratic equation much as I did with plain numbers  $$f = \frac{1}{2} \pm \sqrt{\frac{1}{4} + 1}$$ $$f = \frac{1 \pm \sqrt{1+4}}{2} = \frac{1 \pm \sqrt{5}}{2}\cdot f^0$$  And finally the solution $$f(x) = \frac{1 \pm \sqrt{5}}{2} x .$$  Now my question is: *Is it somehow allowed to work with functions in that way?** I know that in the above, there are denotational ambiguities as $1$ is actually treated as $f^0 = id$ ... But since the result is correct, there seems to be some correct thing in this approach.  So can I actually solve certain functional equations like this? And if true, how would the correct notation of the above be?
51002
Consider the function $f(x,y)=\dfrac{xy(x^2-y^2)}{x^2+y^2}$ for $(x,y) \neq (0,0)$, $f=0$ otherwise.  I have to compute $\dfrac{d^2f}{dydx}(0,0)$.  I know that I have to calculate $\frac{df}{dx}$ first. But that is ,  $\frac{df}{dx} = \frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial x}$. When I put it in wolframalpha, it gives me this calculation, and also just $\frac{\partial f}{\partial x}$ as an alternative form.   Why do I just ignore the y'? I don't know what y(x) is, as a function of x.  y and x are independent functions, no?