7 votes
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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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5 votes
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Solution of double derivative equations

The equation is $$xy''-3y'=4x^2$$ Let $y'=u$ therefore our equation becomes $$u'x-3u=4x^2$$ or $$\frac{du}{dx}x-3u=4x^2$$ Now lets substitute $u=z^2$ therefore $du=2z\:\:dz$ And our equation becomes $$...
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  • 1,413
5 votes
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Implicit Differentiation - $x^my^n = (x+y)^{m+n}$

You just didn't finish. $\begin{array}\\ y' &= \frac{nxy-my^2}{nx^2-mxy}\\ &= \frac{y(nx-my)}{x(nx-my)}\\ &= \frac{y}{x}\\ \end{array} $ Note that if $nx=my$ you can not take this final ...
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5 votes
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Continuity of $f$ doesn't imply differentiability

You are right: from the fact that$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}(x-a)=0,$$you cannot deduce that$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\cdot\lim_{x\to a}(x-a)=0.\tag1\label a$$The equality \eqref{a} ...
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5 votes

How can I argue that $f(x) = (2-x)^3-x+\frac{3}{2}$ is decreasing with a polynomial?

$$f'(x)=-3x^2+12x-13 = -3(x^2-4x+4)-1 = -3(x-2)^2-1<0 \text{ for all } x\in\mathbb R$$
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  • 1,725
4 votes

Is there any meaning for $f(x)-f'(x)$?

If $f$ is a polynomial, then if any of its roots have multiplicity greater than $1$, these roots will be shared between $f$ and $f'$. So, the multiple roots of $f$ are zeros of $f-f'$ with lower ...
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  • 1,225
4 votes
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Rewriting equation for solving differential equation

I'm not exactly sure which part you are struggling with, but my guess is that he is taking some additional steps that you aren't seeing. Also, be sure to note that $\frac{d}{dx}$ is the derivative ...
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  • 2,918
3 votes

Implicit Differentiation - $x^my^n = (x+y)^{m+n}$

Simpler way: Take logarithms (base e unless specified). $$m\log x+n\log y=(m+n)\log (x+y)$$ Now differentiate: $$\frac mx +\frac ny\frac{dy}{dx}=\frac{m+n}{x+y}\left(1+\frac{dy}{dx}\right)$$ Therefore ...
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2 votes

For all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, l(\epsilon)), |f(x+\phi) -f(x)| < \epsilon$

Problem Let $f:\mathbb{R} \rightarrow \mathbb{R}$ differentiable such that $|f'(x)|<K$. Porve for all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, ...
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  • 5,556
2 votes
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Proving strong lyapunov function

From $$\dot{V} = p x \dot{x} + q f(y) \dot{y},$$ the substitutions are made $$\dot{V} = p x \left[- a x + b f(y)\right] + q f(y) \left[c x - d f(y)\right]$$ $$\dot{V} = - a p x^{2} + b p x f(y) + c q ...
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  • 111
2 votes
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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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  • 39.3k
2 votes
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Existence of smooth function with prescribed zeros and value of derivative at the origin

Assume $h$ is $C^1$ and satisfies $h(x, 0) = 0$, and $h(x, f(x)) = 0$ for any $x\in\Bbb R$. Since $h(x, 0) = 0$, and $h(x, f(x)) = 0$ for any $x >0$, there is $0<y_x<f(x)$ such that $\frac{\...
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  • 4,455
2 votes

Differentiating $\sec^{-1}(\sqrt{1+x^2})$ for $x \in (-1, 1)$

I get $${dy\over dx}={1\over1+x^2}$$ for the derivative of $\sec^{-1}(\sqrt{1+x^2})$. I'll put the steps here. $$y=\sec^{-1}(\sqrt{1+x^2})\\ \sec y=\sqrt{1+x^2}\\ {\sin y\over\cos^2 y}{dy\over dx}={x\...
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  • 8,200
2 votes

Derivative of the determinant with respect to the vector

Jacobi's formula for the derivative of the determinant is usually stated as follows: Theorem. For a path $t\mapsto A(t)$ of matrices, one has \begin{equation} \partial_t\det A(t)=\mathrm{Tr}\left(\...
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2 votes

Implicit Differentiation - $x^my^n = (x+y)^{m+n}$

For any homogeneous relation of $x$ and $y$, we always have $$\frac{dy}{dx}=\frac{y}{x}.$$ Because the relationship is always satisfied by $y=vx$ meaning $$\frac{dy}{dx}=v=\frac{y}{x}.$$ The relation $...
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  • 39.3k
2 votes

Taylor Expansion of a function that maps matrices to scalars

Basing on the fact that $\mathbb{R}^{n\times m}\simeq \mathbb{R}^{nm}$ we get for $X=\{x_{ij}\}$ $$\displaylines{f(\tilde{X})\approx f(X)+\sum_{i=1}^n\sum_{j=1}^m{\partial f\over \partial x_{ij}}(X)\,(...
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2 votes

How to solve the following Fourier Transform, knowing the note below?

You could use several properties of the Fourier transform to reduce it to just a few standard transforms. If will assume $$\mathcal F(f)(\xi) = \int_{-\infty}^\infty f(x) e^{-i \xi x}\, dx.$$ ...
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  • 3,338
2 votes
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$f(x):[0,\infty)\to\mathbb R$ is twice differentiable such that $f(0)=0$ and $f''(x)<0$ $\forall x\in(0,\infty)$ and $f(1)=\alpha,\alpha>0.$

For an unbounded above counterexample, we can take: $$f(x)=e^{-a}+Cx-e^{-x-a},\,0\le x$$ For any constants $C>0$, $a>0$. Motivation: although $f’’(x)=-e^{-x-a}<0$, $f’’$ rapidly decays and $...
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  • 12k
2 votes

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, $a_n=\frac{1}n, b_n=\frac{\sqrt{2}}{n},~~ n=1,2,3...$ ...
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  • 5,556
2 votes

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 $x_1 < x_2$ ...
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  • 86.6k
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 ...
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  • 8,193
1 vote
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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 ...
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  • 48.6k
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 ...
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  • 692
1 vote
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an infinitely differentiable monotonic function that goes to 0 at infinity but f' does not

I agree with you, this doesn't seem to work. But if you scale the non-constant intervals in the $x$-direction by a factor of $\alpha$, then you increase $f'(x)$ by a factor of $1/\alpha$. So set $\...
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  • 60.6k
1 vote

For all $\epsilon >0 $, exists $l(\epsilon)$ such that $\forall x \in \mathbb{R}$ and $\phi \in (0, l(\epsilon)), |f(x+\phi) -f(x)| < \epsilon$

Since $f$ is differentiable we have $|f(x+h)-f(x)| = |h||f'(\alpha)|$ for some $\alpha \in (x, x+h)$ by mean-value theorem, which means $|f(x+h)-f(x)| \le K |h|$. For given $\epsilon >0$ let $l(\...
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  • 41
1 vote
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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 ...
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  • 3,298
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. ...
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  • 7,034
1 vote
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Which comes first derivative of force then vector it or derivative of vector force.

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 ...
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  • 5,374
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, $\sec^{-1}(\sec x)$...
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  • 361
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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  • 21

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