OK! Here are some addups to the last part of the first differentiation chapter

by kamatadu, Nov 12, 2022, 10:38 AM

Here are some ideas, which I felt might be worth writing up...

Firstly I was fully confused how to find $\dfrac{d^2 u}{dx ^2}$ where $u$ is a function of $x$. After spending almost $45$ minutes, I finally figured out how to calculate $(du)^{'}_{u}$ which was the remaining part that I was failing to understand how to derive.

Anyways after trying for some time, I figured out how to find $d(u)'_{u}$, and I'm sure ull give me slangs after u notice how dumb of a person I am :)

Firstly, $y = f(u)$ and $u = \phi(x)$. So, after getting value of (du)'_{u}, we continue to solve the main equation below,
\begin{align*} &\dfrac{dy}{du} = \dfrac{d (f(u))}{du} = f'(u)\\ \implies &dy = f'(u) \cdot du. \end{align*}
Now again differentiating both sides,
\begin{align*} \dfrac{d^2 y}{du} &= (f'(u) \cdot du)'_{u}\\ &= f''(u) \cdot du + f'(u) \cdot (du)'_{u}\\ &= f''(u) \cdot du + f'(u) \cdot \dfrac{d^2 u}{du}. \end{align*}
So we finally get, $$d^2 y = f''(u) \cdot du^2 + f'(u) \cdot d^2 u$$:) And we are done lol...

Initially I was failing to understand how do I even find its $du'_{u}$ :huuh: :rotfl: .

O and also one more thing, if $y = f(x)$ (i.e, the variable is now independent), then $dx$ acts more like a constant (cuz now $dx = \Delta x$), so,
\begin{align*} &\dfrac{dy}{dx} = f'(x)\\ \implies &dy = f'(x) \cdot dx. \end{align*}and, \begin{align*} &\dfrac{d^2 y}{dx} = (f'(x) \cdot dx)'_{x}\\ \implies &\dfrac{d^2 y}{dx} = f''(x) \cdot dx\\ \implies &d^2 y = f''(x) \cdot dx^2. \end{align*}
I'll complete this post by adding the leftover screenshots from the book to the IMGUR Link :)

Here it is! IMGUR Link to Image Collection.

Bye again! :P :bye:
This post has been edited 3 times. Last edited by kamatadu, Nov 12, 2022, 11:40 AM
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