8

What is confusing you is that the letters $x$ and $y$ are used for two purposes: (a) As names of the functions $x(t), y(t)$; (b) As names of the "first argument" and "second argument" of $f$. Sadly, this notation is very usual in mathematics, and there is no suitable alternative. If I could convince to write partial derivatives ...


8

I'll spell out @joriki's logic. Define $f(u,\,v):=u^v,\,u:=x,\,v:=x$ so, by the multivariable chain rule,$$\frac{df}{dx}=\color{red}{\frac{\partial f}{\partial u}\frac{du}{dx}}+\color{blue}{\frac{\partial f}{\partial v}\frac{dv}{dx}}=\color{red}{vu^{v-1}}+\color{blue}{u^v\ln u}.$$Alice only used the red term, forgetting $dv/dx\ne0$; Bob only used the blue ...


7

Yes, it's correct. Here's a sketch of the argument. Note first of all that by the implicit function theorem, the portion of a level curve of $F$ in the given region must be a one-dimensional manifold with no boundary points in the open rectangle $(x_1,x_2)\times\Bbb R$. Let $\Gamma$ be the connected component of the level curve $F(x,y)=c$ passing through $(...


6

The standard notation is the $2)$: $$ \frac{\partial^2 f}{\partial x \partial y} := \frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y}\right) $$ EDIT Other notations: $f_{yx}=\frac{\partial^2 f}{\partial x \partial y}$ $\partial_x \partial_y f=\frac{\partial^2 f}{\partial x \partial y} $


6

You have to first compute $\frac {f(h,0)-f(0,0)} h$ and then take the limit. Since $\frac {f(h,0)-f(0,0)} h=0$ for every $h \neq 0$ the limit is $0$. It is not an indeterminate form.


6

If $f$ is differentiable, the directional derivative along $(1,1)$ can be computed as $$ \partial_{(1,1)} f(1,1) = f'_x(1,1) \cdot 1 + f'_y(1,1)\cdot 1. $$ Since the directional derivative is zero ($f$ is constant along that direction), it must be true that $f'_x(1,1) + f'_y(1,1)=0.$


6

All you need to do is differentiate both sides of the symmetry equation- $$(\partial_1 A)(x,m) = \frac d{dx}[A(x,m)] = \frac d{dx}[1-A(m,x)] = - (\partial_2 A)(m,x)$$ or did you want something more?


5

It is defined because the $0$ in the numerator of your limit is "real" $0$. Look, in the expression: $$\lim_{x\rightarrow 0}\frac{0}{x}$$ The answer is in fact $0$, because the denominator will be a very little value close to $0$, but not $0$, while the numerator is just $0$ ($0$ divided by any real number different to $0$ is equal to $0$, and the $...


5

By differentiability theorem if partial derivatives exist and are continuos in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point. The existence of partial derivatives doesn't suffice. In this case we can proceed by definition $$ \lim_{(h,k) \to (0,0)} \frac{\dfrac{2h^2k+k^3}{h^2+k^2}-k}{\sqrt{h^2+k^2}}= ...


5

To complement user's answer, I would like to point out that the example in the OP is even more striking since not only do partial derivatives $\partial_1f(0,0)$ and $\partial_2f(0,0)$ exists, but also the directional derivative of the function $f$ at $\boldsymbol{0}=(0,0)$ along any direction $\mathbf{v}=(h,k)$ exists: $$\partial_\mathbf{v}f(0,0):=\lim_{t\...


5

It's not! The key is local minimum, meaning that $f(1, 1)$ is less than $f(x, y)$ for all $(x, y)$ near $(1, 1)$. More precisely, there is some $r > 0$ such that $$\sqrt{(x - 1)^2 + (y - 1)^2} < r \implies f(x, y) \ge f(1, 1).$$ But, $f(x, y) \ge f(1, 1)$ doesn't hold globally, i.e. $f$ does not have a global minimum at $(1, 1)$. While local minimums ...


5

The function $f$ is not differentiable at $(0,0)$. If it was, then, since $f_x(0,0)=f_y(0,0)=0$, then $f'(0,0)$ would be the null function. In other words,$$\lim_{(x,y)\to(0,0)}\frac{x^2y}{(x^2+y^2)^{3/2}}=0.$$But if $x=y>0$, $\frac{x^2y}{(x^2+y^2)^{3/2}}=\frac1{2\sqrt2}$.


4

The standard notation is to use a hat: writing, for instance, $(a_1,...\widehat{a}_i, …, a_n)$ means that you do not consider the $i$-th term. In this case, however, simply writing: $$ \prod_{1 \leq i \leq n, \, \, i \neq k} x_i ^{\alpha_i} $$ works as well.


4

As noticed in the comments, by chain rule we have for any components $$\frac{\partial}{\partial x_i} \left(\frac1f\right)=-\frac1{f^2}\frac{\partial f}{\partial x_i}$$ and therefore $\nabla\left(\frac{1}{f}\right)=-\frac{1}{f^2}\nabla f$. Let consider as a concrete example $f(x,y)={x^2+y}$ then $$\frac{\partial}{\partial x} \left(\frac1{x^2+y}\right)=-\frac{...


4

It is easier if you substitute $x$ and $y$ directly $$g(r,\theta)=\frac{1}{x+y^2}=\frac{1}{32r\cos(\theta)+9r^2\sin^2(\theta)}$$ So $$\frac{\partial g}{\partial \theta}=-\frac{-32r\sin(\theta)+18r^2\sin(\theta)\cos(\theta)}{(32r\cos(\theta)+9r^2\sin^2(\theta))^2}.$$ Or $$\frac{\partial g}{\partial \theta}=\frac{\partial g }{\partial x}\frac{\partial x}{\...


4

Since the objective function is a function of $y,z$ can we rewrite the constraints to eliminate x, and have objective and constraints in the same variables? $3x + z = 5\\ x = \frac {5-z}{3}\\ x^2 + y^2 = 1\\ \left(\frac {5-z}{3}\right)^2 + y^2 = 1$ We have an ellipse. We need to find where the tangent of the ellipse is parallel to $y+4z$ $-2\frac {5-z}{9}\ ...


4

First of all, the idea behind the operator being multiplied on the right by a matrix is that it will still output an operator. For example, operating on a function $f$ we would find $$\begin{aligned} \left(\begin{array}{c} \frac{\partial}{\partial r} \\ \frac{\partial}{\partial \theta} \end{array}\right)[f]=\left(\begin{array}{ll} \frac{\partial x}{\partial ...


4

Your computation is correct (although at the very beginning I would write $d/dx$, since your contour integral is a function of $x$ only). You need to think of $\gamma_x$ as a variational vector field along the curve $\Gamma_x = \partial\Omega(x)$ and then the second integral is a contour integral over $\Gamma_x$ as well. EDIT: In particular, we have the ...


4

I suggest using the definition of partial derivative to have a clearer point of view on the situation. The partial derivative of $A$ with respect to the first variable at the point $(x_0,m_0)$ is defined by : $$\partial_1 A(x_0,m_0)=\lim_{h\to0} \frac{A(x_0+h,m_0)-A(x_0,m_0)}{h}.$$ Using the symmetry relation, we have : $$\lim_{h\to0} \frac{(1-A(m_0,x_0+h))-(...


4

In the context of the chain rule, it is understood as you said (and as I said in the comments), in the way that you described as "magically". Let me give an example to show a slightly different perspective, which hopefully will be the more "natural" interpretation you are looking for. Consider a circle, with radius $r$. Its area is $A = \...


4

Let's call $A=X\Lambda X^T$. Now, $$\frac{\partial}{\partial \alpha_i}A^{-1} = -A^{-1} \left( \frac{\partial}{\partial \alpha_i} A \right)A^{-1}$$ (see Derivative of the inverse of a matrix), provided $A^{-1}$ exists. Finally, $$\frac{\partial}{\partial \alpha_i} A= x_i x_i^T$$.


3

The KP equation was derived from physical principles, so you should be able to find this derivation somewhere. Worst case scenario, you may have to refer to the original paper of Kadomtsev and Petviashvili, who derived it for plasma waves; however, I believe that paper is in Russian. And it's probably not helpful if you aren't familiar with plasma physics. ...


3

I'm not sure how one would derive the analog, but my gut feeling is that one would generalize the KdV hamiltonian (which is the second (aka "energy") integral of motion on the wikipedia page). I'm not sure if Hamiltonian mechanics is something you've covered in engineering courses, as it can quickly turn into very advanced/technical pure ...


3

If you are familiar with the language of differential forms, the Laplacian is $$\star d \star d,$$ where $\star$ is the Hodge star operator and $d$ is exterior derivative. In polar coordinates, the volume form is $r dr \wedge d\theta$. The Hodge star operator sends the volume form to $1$ ($\star r dr \wedge d\theta = 1$) and acts on the $1$-form basis as ...


3

What you have is a transport equation. Observe that $\partial_1 f-\partial_2 f=\nabla f\cdot (1,-1)$ is just the directional derivative of $f$ in the direction $(1,-1)$. Your PDE means that your function is a constant as you move in this direction. But moving in this direction starting at some point on the axis $t=0$ takes you to any point in $\mathbb{R}^2$,...


3

The fact that $f(x-y),f(x),f(y),f(x+y)$ is an Arithmetic Progression implies that $$f(x-y)+f(x+y)=f(x)+f(y)$$ Substituting $y=0$ gives us $$2f(x)=f(x)+f(0)\to f(x)=f(0)$$This tells us that the function is constant, i.e., $f'(x)=0$, so it is even and odd. Your work is perfectly fine. Your last equation implies that $f'(x)=0$, but I think the proof I've ...


3

It looks good to me. Just a few comments and an alternative easier solution. If it was the case, the following limits should be equal due to the property of composition of limits: \begin{align*} \lim_{t\to 0}f(\Gamma_{1}(t)) = \lim_{t\to 0}f(\Gamma_{2}(t)) \end{align*} whenever $\Gamma_{1}(t)\to(0,0)$ and $\Gamma_{2}(t)\to(0,0)$ as $t\to 0$. Having ...


3

Recall from the Fundamental Theorem of Calculus, if we define ${f(x)=\int_{a}^{x}g(t)dx}$ then we have ${f'(x)=g(x)}$. In your case, a multivariate function is defined to be $${f(x,y)=\int_{x}^{y}e^{-t^2}dt}$$ Remember when computing the partial derivatives of ${f(x,y)}$, you always treat the other variable as a constant. So in actuality we get $${\frac{\...


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