I have a function $f(x)$ defined as follows: $$ f(x)=\frac{1-g(x)}{1-xg(x)}. $$ Because $f$ contains the function $g(x)$, I guess you could say $f$ is a function of $g(x)$ and $x$. Given this, a mathematician says $$ \frac{\partial}{\partial x}f=\frac{\partial f}{\partial g}\frac{\partial g}{\partial x} + \frac{\partial f}{\partial x}. $$ I’m not following his reasoning. For one, I […]

This previous question of mine has lead me to ask the following question: It was my understanding that the chain rule $$\dfrac{du}{dx}=\dfrac{dy}{dx}\dfrac{du}{dy}$$ only makes sense when there is some function $u$ for it to operate on. So how can we possibly justify writing $$\dfrac{d}{dx}=\dfrac{dy}{dx}\dfrac{d}{dy}?$$ One of the answers to the previous question mentioned that if […]

I saw this proof of the chain rule but it says this is a flawed proof. Why? I guessed the reason it is wrong because you can’t substitute $g(x+h)$ and $g(x)$ into in limit.

I have just begun studying the Calculus of Variations and I have 3 doubts in it. I have written certain things in bold so as anyone who wishes to answer the question but find it quite long can just go through the bold part and he would get an idea as what my doubt is […]

Show that the time-independent Schrödinger equation for a simple harmonic oscillating potential $$-\frac{\hbar^2}{2m}\frac{d^2 u}{dx^2}+\frac12 m\,\omega_0^2 x^2u=E\,u$$ can be written as $$\frac{d^2u}{dy^2}+(\alpha – y^2)u=0\tag{1}$$ where $$y=\sqrt{\frac{m\,\omega_0}{\hbar}}x$$ and $$\alpha=\frac{2E}{\hbar\,\omega_0}$$ So by my logic $$\frac{dy}{dx}=\sqrt{\frac{m\,\omega_0}{\hbar}}$$ and $$\frac{d^2y}{dx^2}=0$$ clearly something has gone wrong or I am going about this the wrong way. My lecturer mentioned in the lecture that: […]

In Leibniz notation of the chain rule, $$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$$ Where $y\left ( u\left ( x \right ) \right )$ is a composite function of x. I understand that the du‘s don’t simply cancel out because $\frac{dy}{du}$ and $\frac{du}{dx}$ are defined as specific limits making the numerator and denominator infinitesimals and thus making […]

I know that if $u=u(s,t)$ and $s=s(x,y)$ and $t=t(x,y)$ then the chain rule is $$\begin{align}\color{blue}{\fbox{$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial s}\times \frac{\partial s}{\partial x}+\frac{\partial u}{\partial t}\times \frac{\partial t}{\partial x}$}}\color{#F80}{\tag{A}}\end{align}$$ A short extract from my book tells me that: If $u=(x^2+2y)^2 + 4$ and $p=x^2 + 2y$ then $u=p^2 + 4$ therefore $$\frac{\partial u}{\partial x}=\frac{\partial u}{\partial p}\times \frac{\partial […]

When deriving $x^x$, why can’t you choose $u$ to be $x$, and find $\dfrac{d(x^u)}{du} \dfrac{du}{dx} = x^x$? Or you could go the other way and find $\dfrac{d(u^x)}{du}\dfrac{du}{dx}$, giving $\ln(x)\cdot{x^x}$? Both methods seem to be equally wrong.

This is what I worked out: Let $y = (1 + x^2)^x$ and let $a = 1 + x^2$ Then, by the chain rule of differentiation: $\frac{dy}{dx} = \frac{dy}{da}\cdot\frac{da}{dx} = a^x\cdot ln(a) \cdot 2x$ $\frac{dy}{dx} = (1 + x^2)^x \cdot ln(1 + x^2) \cdot 2x $ But when I try to verify the result on […]

This question is a simplified version of this previous question asked by myself. The following is a short extract from a book I am reading: If $u=(x^2+2y)^2 + 4$ and $p=x^2 + 2y$ $\space$ then $u=p^2 + 4=f(p)$ therefore $$\frac{\partial u}{\partial x}=\frac{\rm d f(p)}{\rm d p}\times \frac{\partial p}{\partial x}=2xf^{\prime}(p)\tag{1}$$ and $$\frac{\partial u}{\partial y}=\frac{\rm d f(p)}{\rm […]

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