# Tag Info

## New answers tagged derivatives

0

You have gone astray in several ways here. What you are looking for is points where both $\frac {\partial f}{\partial x} = 0$ and $\frac {\partial f}{\partial y} = 0$, but you've gotten distracted by where $\frac {\partial f}{\partial x} = \frac {\partial f}{\partial y}$, which indeed is necessary, but is not sufficient. The set of equations you need to ...

1

No. $$[\sin^2(x)\cos(x^2)(2x)] + \sin(x^2)(2\sin x \cos x)$$ can be rewritten as $$2\sin x \left(x\sin(x) \cos(x^2)+\sin(x^2)\cos(x)\right)$$

3

We get \begin{align*} &(\sin^2 x)(\cos(x^2)) (2x)+(\sin (x^2))(2\sin x\cos x)\\ &=2\sin x(x\sin x\cos(x^2)+\sin(x^2)\cos x) \end{align*} simply by factoring the $2\sin x$ from each term. This is a bit different than the answer you provided, but it is what the answer should be.

1

I get \begin{align*} \frac{d}{dx}\left[\sin^2(x)\sin(x^2)\right]&=2\sin(x)\cos(x)\sin\left(x^2\right)+\sin^2(x)\cos\left(x^2\right)(2x)\\ &=2\sin(x)\left[\cos(x)\sin\left(x^2\right)+x\sin(x)\cos\left(x^2\right)\right]. \end{align*} The term with $\cos\left(x^2\right)$ should still have a $\sin(x)$ multiplying it after factoring, because it was ...

1

The way your Poisson Bracket is written, it does not make sense that $f$ and $g$ are "vector valued". So I will assume scalar values, with vector inputs. $$f,g:\mathbb{R}^n\mapsto \mathbb{R}$$ Using index notation you have \begin{align} \frac{\partial f(x+Ay)}{\partial y_i} &= \frac{\partial f(x+Ay)}{\partial x_j}\frac{\partial (Ay)_j}{\partial y_i} ... 1 Theorem 25.1 in Rockafellar's Convex Analysis states that if the subdifferential of f at x is a singleton, then f is differentiable at x. Let u,v\in \partial f(x). Note that for any direction y, if \alpha >0\langle y,u \rangle = \frac 1{\alpha} \langle y,\alpha u \rangle \leq \frac{f(x+\alpha y)-f(x)}{\alpha}$$and if \alpha <0$$\...

0

Here is a sketch of the missing algebra. Given the FOC in terms of the row vector $a^T$ \eqalign{ (x^Tx)a^T + \Big(\frac{\lambda}{\sqrt{a^Ta}}\Big) a^T = x^TR \\ \Bigg(x^Tx + \frac{\lambda}{\sqrt{a^Ta}}\Bigg) a^T = r^T \\ \Bigg(\chi^2 + \frac{\lambda}{\alpha}\Bigg) a^T = r^T \\ } So $a$ is seen to be a scalar multiple of $r$. Square both sides and solve ...

1

An alternative approach is to use a Taylor expansion approach $$f(X + \varepsilon M) = tr( Y (X + \varepsilon M)^{-1}) = tr( Y X^{-1} (I + \varepsilon M X^{-1})^{-1})$$ and $$(I + \varepsilon M X^{-1})^{-1} = I - \varepsilon M X^{-1} + O(\varepsilon^2)$$ which yields $$f(X + \varepsilon M) = f(X) - \varepsilon tr( Y X^{-1} M X^{-1} ) + O(\varepsilon^2) ... 1 Apply the first part of the fundamental theorem of calculus$$\frac{d}{dt}\int_a^t f(x)dx=f(t)$$where every x in the integrand is replaced by t. 3 You seem to be thinking of {\mathrm{d}\over\mathrm{dx}} as one of the functions in the chain rule. The chain rule (in one variable) applies to the composition of two real-valued functions of a real variable. {\mathrm{d}\over\mathrm{dx}} is not such a function. It takes differentiable functions of a real variable to functions of a real variable. The ... 2 The basic thing you need is f_u and f_v. I'll describe how to get the first. The second is like unto it. So, we have that$$f_u=\frac{f_x}{u_x}+\frac{f_y}{u_y}.$$Similarly, obtain f_v. Then the gradient is (f_u,f_v). Thus, at the point (x,y)=(-1,2), the gradient is given by \nabla_{(-1,2)}=(f_u(-1,2),f_v(-1,2)), so that in the direction (3,-4),... 3 This is false as stated (probably it's true with some reasonable additional hypothesis, maybe regarding monotonicity of f'). The Idea: We take f(x)=\int_0^x f'(t)\,dt, where f'>0 is constructed so that f'(n)\to\infty, but f' is very small except at points very close to a positive integer, so that \int_0^\infty f'\le1. Then f(x)\le1, so the ... 2 Because f is differentiable at 0 (note that we do not even need that f is differentiable or even continuous elsewhere), we have that \lim_x\to 0{f(x)-f(0)}{x} exists, hence with \epsilon:=1 there exists \delta>0 such that$$\left|\frac{f(x)-f(0)}{x}-f'(0)\right|<1 $$for |x|<\delta. In paticular, f(x) is between two lines with ... 4 In order for f to be differentiable in 0, it has to be continuous in 0. Therefore$$\lim_{n\rightarrow\infty}f(x_{n})=f(0)$$if \lim_{n\rightarrow\infty}x_{n}=0. 0 Hint An easier solution is$$f(x)=e^{ax}\sin (bx+c)=\Im\{ e^{ax+jbx+jc}\}$$therefore$${d^n f(x)\over dx^n}=\Im\{{d^n\over dx^n} e^{(a+jb)x+jc}\}=\Im\{e^{jc}{d^n\over dx^n}e^{(a+jb)x}\}=\Im \{e^{jc}(a+jb)^ne^{(a+jb)x}\}$$0 The function x\mapsto|x| is differentiable over \mathbb{R}\setminus\{0\}, so there is no problem in applying the chain rule: for x\ne0,$$ \frac{d}{dx}\log\lvert x\rvert=\frac{1}{\lvert x\rvert}\frac{\lvert x\rvert}{x}=\frac{1}{x} $$0 First things first, no expression will work at x=0, since \log|x| is not defined there. Also, |x| is differentiable everywhere except at x=0, so things should work fin elsewhere. It's often easier to work with |x| as a piecewise-defined function, so: If x>0, |x|=x, you can apply the chain rule there and get f'(x)=\frac{1}{|x|}=\frac{1}{x} ... 0 For non-zero x, we could still use the formula \frac{d}{dx}\log f(x)=\frac{f'(x)}{f(x)} Since f'(x)=\begin{cases}1&x>0\\-1&x<0\end{cases} for f(x)=|x| so \frac{d}{dx}\log |x|=\begin{cases}\frac{1}{|x|}&x>0\\\frac{-1}{|x|}&x<0\end{cases}=\frac{1}{x} given x\neq 0 4 Since \log(|x|) is not defined at x=0, we may investigate the differentiability for x\not=0. We distinguish two cases depending on the sign of x. For x>0, we know that |x|=x and$$\frac{d}{dx}\left(\log(|x|)\right)=\frac{d}{dx}\left(\log(x)\right)=\frac{1}{x}.$$On the other hand, for x<0, we have that |x|=-x and therefore$$\frac{d}{...

1

$\frac d {dx} \log|x|=\frac 1 x$ if $x \neq0$ but the derivative at $0$ does not even makes sense since the function is not defined at $0$.

1

What you have done is correct. The only possible solution is $f(x)=\frac 1 {g(x)} {\int_0^{x} yg'(y)dy}$ for $x>0$. But now $f(x)\to 0$ as $x\to 0$ at least when $g$ is a nice function so the given IVP has no solution. [ $|f(x)| \leq \frac {x\int_0^{x} g'(y)dy} {|g(x)|}= x \to 0$ if $g'$ is positive, for example].

1

First, note that the title in the excerpt is incorrect. It should be "$n$th derivative of $e^{ax}\sin(bx+c)$". (1) Here we introducing new quantities $r$ and $\alpha$, so we may define them however we want. Essentially, this amounts to writing the pair $(a, b)$ in polar coordinates, as $(r, \alpha)_{\textrm{polar}}$. More explicitly, any such $r$ satisfies ...

1

$$y=~e^{ax}\sin(bx+c)~$$ Differentiating with respect to $~x~$, we have $$y_1=\frac{dy}{dx}=~a~e^{ax}\sin(bx+c)~+~b~e^{ax}\cos(bx+c)~$$ $$\implies y_1=~e^{ax}~\{~a~\sin(bx+c)~+~b~\cos(bx+c)\}~\tag1$$ For computation of higher-order derivatives it is convenient to express the constants $~a~$ and $~b~$ in terms of the constants $~r~$ and $~\alpha~$ defined by ...

1

Differentiate $f\circ \gamma$ (I use primes instead of overdots): $$(f\circ \gamma)'=[f( \gamma(t))]'=f'( \gamma)\gamma'(t)=(f'\circ \gamma)\gamma'.$$ Thus, we have that $$|(f\circ \gamma)'|=|f'( \gamma)||\gamma'(t)|=(r\circ\gamma)|\gamma'|=(r\circ\gamma)\gamma',$$ if $\gamma'\ge 0.$ Also, we have that \frac{(f\circ \gamma)'}{|(f\circ \gamma)'|}=\frac{(f'... 2 This is more or less how the proof of the power rule goes. The binomial theorem is handy to generalise the idea to arbitrary integral indices, but the proof of a specific case is much more clear to follow at first: \begin{align*} y + \delta y &= (x+\delta x)^3\\ &= x^3 + 3x^2\delta x+3x\delta x^2 + \delta x^3\\ \implies \delta y &= \... 1 Let f(\mathbf{x})=\mathbf{x}^\top\mathbf{x}. If write \mathbf{x}=\mathbf{x}_0+\epsilon \mathbf{y} in the neighborhood of a point \mathbf{x}_0 then we may write \begin{align} f(\mathbf{x}) &=\mathbf{x}^\top\mathbf{x}\\ &=(\mathbf{x}_0+\epsilon \mathbf{y})^\top (\mathbf{x}_0+\epsilon \mathbf{y})\\ &=\mathbf{x}_0^\top \mathbf{x}_0+(2\mathbf{x}... 3 Consider the trace/Frobenius product (denoted by a colon)A:B = {\rm Tr}(A^TB)$$As long as the matrices (A,B) have same number of rows/columns, they can have any shape: tall-and-thin, square, short-and-fat. And of course, they can be row or column vectors in which case one recovers the ordinary dot-product$$a:b = {\rm Tr}(a^Tb) = a\cdot b$$The ... 0 If this is related to the differential operator d from differential geometry, then a property of d is that d(dx) = 0 and d(uv) = (du)v + u(dv). 0 You seem to be on the right path to show$$ \mathrm{d}\left( \frac{x \,\mathrm{d} x+y \,\mathrm{d}y}{\sqrt{x^2 + y^2}} \right) = \frac{\left(x^2+y^2\right) (x \,\mathrm{d}^2 x+y \,\mathrm{d}^2 y)+x^2 (\mathrm{d}y)^2+y^2 (\mathrm{d}x)^2-2 x y \,\mathrm{d}x \,\mathrm{d}y}{\left(x^2+y^2\right)^{3/2}} \text{.} $$Then$$ \frac{(y \,\mathrm{d}x - x \,\...

2

I think you are assuming that $\sum_{i=1}^n\sqrt{(x-a_i)^2 + (y-b_i)^2} = R \space \Rightarrow \space \sum_{i=1}^n(x-a_i)^2 + (y-b_i)^2 = R^2$ which is incorrect (unless $n=1$).

3

In order to find $$d(xdx)$$ you apply the product rule of differentiation. $$d(xdx)= dxdx + xd^2x = (dx)^2 + xd^2x$$

3

As $x^Tx$ is a scalar, $\color{blue}x^Tx=(\color{blue}x^Tx)^T=x^T\color{blue}x.$

2

You can take the differential of the relationship \eqalign{ p &= (1+r)(pA+wl) \\ } to obtain (assuming $p,r$ are constant) \eqalign{ 0 &= (1+r)(p\,dA + l\,dw+ w\,dl) \\ l\,dw &= -(p\,dA + w\,dl) \\ } This indicates how $(A,l)$ would need to change to compensate for a change in $w$, in such a way that $p$ is held constant. For example, ...

0

You know that $f'(z)=$ $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right]$ then $f'(a,b)$ equal $\left[\begin{array}{cc} u_x & -v_x \\ v_x & u_x \end{array}\right] \left[\begin{array}{c} a\\ b \end{array}\right]=\left[\begin{array}{c} au_x -bv_x\\ av_x+bu_x \end{array}\right]$ The book consider $a+ib = \left[\begin{... 0 As to the title of your question, Logistic function: where does it come from?, I can provide an intuition for the logistic function which is the common interpretation from a machine learning perspective. It seems that the underlying question has already been answered above, but I thought this interpretation could help with your intuition about logistic ... 6 If$f$is differentiable and$f(x_1) = f(x_2) = f(x_3) = 0$with$x_1 < x_2 < x_3$then you can apply the mean-value theorem (or Rolle's theorem) to both intervals$[x_1,x_2]$and$[x_2, x_3]$. It follows that$f'$has a root in each of the open intervals$(x_1, x_2)$and$(x_2, x_3)$. That makes (at least) two roots of the derivative. In the same ... 2 I think you can find all ingredients in Ullrich's book on pages 4-6. Ullrich explains that$f$is complex differentiable at$z$iff there exists$a \in \mathbb C$such that$f(z+h) = f(z) +a \cdot h + o(h)$. Then$f'(z) = a$. The map$L : \mathbb C \to \mathbb C, L(h) = a \cdot h = f'(z)\cdot h,$is$\mathbb C$-linear and thus trivially also$\mathbb R$-... 3 Writing your limits with an$h$makes both equations look like the same limiting procedure. But there is a subtle difference: in your first equation we have$h \in \mathbb{C}$, whereas the second equation is real and hence$h \in \mathbb{R}$. Consequently the first equation is a lot stronger than the second. Instead of approaching$0$only from the left or ... 2 Subtract$x_t$to the LHS and RHS of [A] : $$x_{t+1}-x_t=k x_t (1-x_t)- k \frac{1}{k}x_t$$ $$\underbrace{\dfrac{x_{t+1}-x_t}{1}}_{\text{Discrete derivative}}=k x_t(1 - L x_t) \ \ \text{with} \ \ L:=1+\frac{1}{k}$$ Or, better, under the form (thanks to @Yuriy S for this remark) : $$\underbrace{\dfrac{x_{t+1}-x_t}{\Delta t}}_{\text{Discrete derivative}}=k'... 0 First, let's discuss what this equation means.$$f''(x)g'(x) + g''(x)f(x) = 0$$It is by itself a relation between the two functions. If we fix f(x) and two initial (or boundary) conditions, we can use the ODE to find g(x). On the other hand, if we fix g(x) and two initial (or boundary) conditions, we can use the ODE to find f(x). I doubt we can ... 1 Let h=g'. Then \frac {h '} h=-\frac {f''} f. So log (h(x))=C-\int_0^{x} \frac {f''(t)} {f(t)}\, dt. Take exponential and integrate again to write g in terms of f. 1 The angle of rotation of the clock's minute hand is 2\pi for 60 minutes. For one minute the angle of rotation is \frac{2\pi}{60}=\frac{\pi}{30}. So the tip of the hand travels \frac{\pi}{30}r on the circle. This is a short distance, compared to the circumference of the circle. So the small arc of circle can be confused with a straight line of same ... 0 If we measure from the center of the clock, at 6:01 the x position of the hand is r\cos \frac \pi {30} and at 6:02 it is r \cos \frac \pi{15}. If the time is t minutes past 6:01 the linear approximation to the horizontal position is r\cos \frac \pi {30}+t(r \cos \frac \pi{15}-r \cos \frac \pi{30}). It sounds like you are asked for one value,... 1 There is no difference between the notations - they mean exactly the same thing. However, at different times you will find one more useful than the other. For example, when doing u-substitution with integrals, the \frac{d}{dx} is helpful. The same thing is true when using the chain rule - it is often easier to keep track of what is happening with \frac{d}{... 0 Using prime ' usually indicates that we take the total derivative w.r.t. to all variables, whereas \frac{d}{dx} indicates that we take the total derivative with respect to the variable x. Note that this is different from the partial derivative \frac{\partial }{\partial x}. As an example, consider f(x, y, z)=x^2+y^2+3z where y=\sin(x). Then f' = ... 0 There is no mathematical difference. They are different notations for the same thing. (The first was due to Newton, the second to Leibniz.) 0 To deal with the x^3, observe that you can apply the geometric series trick$$\frac{1}{8-x^3}=\frac{1}{8}\frac{1}{1-\frac{x^3}{8}}=\frac{1}{8}\Big(1+\frac{x^3}{8}+\frac{x^6}{64}+\frac{x^9}{512}+\dots\Big)$$and manipulate the Maclaurin Expansion (a Taylor series expansion centered at x=0). The Maclaurin Expansion of \sin(x) is$$\sin(x)=\sum_{k=0}^{\... 0$0=f'(0)=a(a-1)(a+2)\implies a=0,1$or$-2$. Meanwhile,$0\le f''(0)=2a\implies a\ge0$. Hence$a=0,1$. 0 Just wanted to cover up the$\implies$case, where the other answers doesn't addressed, and the OP just explained vaguely "differentiating and using the chain rule gives that the required derivatives of$f$vanish". As stated in a similar question, it is not just that simple. Let's show that if$f \circ \psi^{-1}\colon \widetilde{\mathbb{R}}\to \mathbb{R}$... 0 If you want to refer to Wikipedia, you have to identify the proper variables. Here, you have$z(y)=f(y)$and$y(x,s)=x+s$. Hence$\frac{dz}{dy}= \frac{df}{dy}$and$\frac{\partial y}{\partial s}= 1\$. Which leads to $$\frac{\partial f(x+s)}{\partial s}= \frac{df}{dy}(x+s)= f^\prime(x+s).$$

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