Derivating $\frac {d^2y}{dt^2}$ using $x,y,t$

Question

I read Boyce, Diprima 7th ed. without knowing anything about relationship between $x,y,t$ we want to derive $d^2y/dt^2$. It is basically chain rule and I could derive by using closed functional forms as following

$$(f(g(h)))'=f'(g(h))g'(h)\\ \; \\\Rightarrow (f(g(h)))''=(f'(g(h)))'g'(h)+g''(h)f'(g(h))$$

$(f'(g(h)))'=f''(g(h))g'(h)$

$$(f(g(h)))''=f''(g(h))g'(h)^2+g''(h)f'(g(h))$$

, but I could not derive as following:

$$\dfrac{dy } {dt }=\dfrac{d y } {d x }\dfrac{d x} {d t }\\\;\\\Rightarrow \dfrac{d^2y } {dt ^2 }=\dfrac{d } {d t}\left(\dfrac{d y} {d t}\right)=\dfrac{d } {d t}\left( \dfrac{d y } {d x }\dfrac{d x} {d t }\right)\\\;\\=\underbrace{\dfrac{d } {d t}\left(\dfrac{d y } {d x }\right)}_{problematic}\dfrac{d x } {dt }+\underbrace{\dfrac{d^2 x} {d t^2 }\dfrac{d y } {dx }}_{okey}$$

I couldnot derive left part because of $\dfrac{d } {d t}\left(\dfrac{d y } {d x }\right)$ everytime I try I find identically(when I add the whole thing I go back $\dfrac{d^2 y } {d t^2}$) not to what I want.

Attempt :

$\dfrac{d } {d t}\left(\dfrac{d y } {d x }\right)=\dfrac{d } {d t}\left(\dfrac{d y/dt } {d x/dt }\right)=\dfrac{y''x'-x''y'}{(x')^2}=y''/x'-x''y'/(x')^2$ not good .

Where is the problem, when I use functional form I can easily derive, but why cant I do it with second method?

Answer 1

Use the same idea like in the first derivative

$$\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)\cdot \dfrac{dx}{dt}\implies \dfrac{d\left(\dfrac{dy}{dx}\right)}{dt}\cdot \dfrac{dx}{dt}\implies \left[\dfrac{d\left(\dfrac{dy}{dx}\right)}{dx}\dfrac{dx}{dt}\right]\cdot \dfrac{dx}{dt} \\\implies\dfrac{d^2y}{dx^2}\cdot\left(\dfrac{dx}{dt}\right)^2$$

Answer 2

Maybe I understand the question wrong, but: $$\frac {d^2y}{dt^2} = \frac {d^2}{dt^2}y \implies y''(t)$$

Answer 3

Not sure to understand the question but for the problematic term apply the chain rule again

$$\dfrac{dy } {dt }=\dfrac{d y } {d x }\dfrac{d x} {d t }\\\;\\\Rightarrow \dfrac{d^2y } {dt ^2 }=\dfrac{d } {d t}\left(\dfrac{d y} {d t}\right)=\dfrac{d } {d t}\left( \dfrac{d y } {d x }\dfrac{d x} {d t }\right)\\\;\\=\underbrace{\dfrac{d } {d t}\left(\dfrac{d y } {d x }\right)}_{problematic}\dfrac{d x } {dt }+\underbrace{\dfrac{d^2 x} {d t^2 }\dfrac{d y } {dx }}_{okey}$$ $$\dfrac{d^2y } {dt^2 }=\left (\frac {dx}{dt} \right)^2 \frac {d^2y}{dx^2}+\dfrac{d^2 x} {d t^2 }\dfrac{d y } {dx }$$