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### Are there any non-constant differentiable functions $f : \mathbb R \to \mathbb R$ where each $t \in \mathbb R$ has $f(t)f(f'(t))=1$?

The property that every $t\in\mathbb{R}$ satisfies \begin{equation}\tag{1}\label{functional-identity} f(t)f(f'(t))=1 \end{equation} is very restrictive. In particular, we can deduce uniqueness of ...
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Accepted

### Finding the maximum distance from origin of a point on the curve $x=a\sin t-b\sin\left(\frac{at}b\right), y=a\cos t-b\cos\left(\frac{at}b\right)$

Let $D(t)$ be the distance of point P from origi On. Then $$OP=D(t)=\sqrt{a^2+b^2-2ab\cos c t}, c=(b-a)/a,~ a,b>0$$ $D(t)$ will admit maximum value when $ct=\pi$, hence $D_{max}=a+b.$
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### Prove that $f(x)$ is not differentiable at $0$

By definition, $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}$$ $0$ is rational, so $f(0)=0$ $$f'(0)=\lim_{x\to0}\frac{f(x)}{x}$$ Consider two sequences, $a_n=\frac{1}n, b_n=\frac{\sqrt{2}}{n},~~ n=1,2,3...$ ...

### How is it that the second derivative of a function can be 0 at a maximum?

Consider the function $g(x) = 4x^3.$ This is an increasing function. It is increasing everywhere you look. No matter what values of $x_1$ and $x_2$ you choose, it is always true that if $x_1 < x_2$ ...
1 vote

### Does Left Hand Derivate and Right Hand Derivative being defined guarantee continuity?

Let $L^{+}$ denote the right hand limit. For $h > 0$ we have $f(a + h) = f(a) + hL^+ + o(h)$. Clearly $hL^+ + o(h) \to 0$ as $h \to 0$, so $f$ is right-continuous at $a$. The same argument shows ...
1 vote
Accepted

### How to comput Hessian or second derivative using chain rule.

We want to compute the gradient and Hessian of the function $L(x) = f(g(x))$, where $g(x) = Wx$. The derivative of $g$ is $g'(x) = W$. By the chain rule, $$L'(x) = f'(g(x)) g'(x) = f'(Wx) W.$$ Note ...
1 vote

### Find all $f:\mathbb{R}\to \mathbb{R}, f\in C^1$ so that $q, f(q)$ has the same denominator as $q$

Suppose $f\left(\frac{an+1}{bn}\right) = f\left(\frac{a}{b}\right)$. Because the denominator of $q$ and $f(q)$ is equal in lowest terms, and we know $\gcd(a, b) = 1$ that implies $\frac{a}{b}$ is in ...
1 vote
Accepted

1 vote
Accepted

### Convergence of derivatives near the boundary of an open interval

There is a subtle but important difference between the title of your question and the body of the post. Your title (and last sentence in the post) talk about convergence, yet the first part of your ...
1 vote
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### continuity of incremental ratio (different quotient)

Consider $A := \{(x,y) \in (a,b)^2 : x < y\}$. Then the function $$g: A \to \mathbb{R}, \quad g(x,y) = \frac{f(x)-f(y)}{x-y}$$ is continuous. As $A$ is connected, $g(A)$ is necessarily an interval. ...
1 vote
Accepted

Suppose for example that one point is not moving, and the other point is moving in a circle around the first. Then $r = |\vec r|$ is constant, so $dr/dt = 0$, but $\vec F$ is not constant, so $d \vec ... 1 vote ### Differentiating$\sec^{-1}(\sqrt{1+x^2})$for$x \in (-1, 1)$Let$\phi = \sec^{-1}(\sec \theta)$. Then$\sec(\phi)=\sec\big(\sec^{-1}(\sec \theta)\big)= \sec(\theta)$, because$\sec(\sec^{-1} x)=x$is always true (although, as you point out,$\sec^{-1}(\sec x)$... 1 vote ### Differentiating$\sec^{-1}(\sqrt{1+x^2})$for$x \in (-1, 1)$The easiest way is through implicit differentiation. Suppose that$y=\sec^{-1}{\sqrt{1+x^{2}}}$This implies that$\sec{y}=\sqrt{1+x^{2}}$Taking the derivative with respect to$x$, you get:$\sec{y}\...

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