Given $x=lnt$ how to find $\frac{d^2y}{dt^2}$ in terms of $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$

Question

I got as far as finding: $$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$

But when taking the second derivative I end up with the following mess:

$$\frac{d^2y}{dt^2}=\frac{d\frac{dy}{dx}}{dt}\frac{dx}{dt}+\frac{dy}{dx}\frac{d^2x}{dt^2}$$

Is there a way to simplify $\frac{d\frac{dy}{dx}}{dt}$ or am I making a mistake in my application of the derivative?

Answer 1

I'm going to give a proper answer because this question has been bothering me as well and it took me a bit of thinking to resolve it. And yeah, like you said, the page I linked is confusing, it goes into far too much depth

Simplifying $\frac{d\frac{dy}{dx}}{dt}$ goes as follows:

$\frac{d}{dt}(\frac{dy}{dx})=$ $\frac{dx}{dt}\frac{d}{dx}(\frac{dy}{dx})=$ $\frac{dx}{dt}\frac{d^2y}{dx^2}$

So you end up with the second total derivative being:

$\frac{d^2y}{dt^2}=\frac{d^2y}{dx^2}(\frac{dx}{dt})^2+\frac{dy}{dx}\frac{d^2x}{dt^2}$

Answer 2

Let $x=g(t), y=f(x)$, so that $y=(f\circ g)(t)$.

Using Newton's dash notation for derivation, we have:

$\newcommand{\de}{\operatorname d}\begin{align} \frac{\de^2 y }{\de t\;^2} & = (f\circ g)''(t) \\[1ex] & = (g' \cdot f'\circ g)' (t) \\[1ex] & = ((g')' \cdot f'\circ g)(t) + (g'\cdot (f'\circ g)')(t) \\[1ex] & = (g'' \cdot f'\circ g)(t) + (g'^2\cdot f''\circ g)(t) \\[1ex] & = \dfrac{\de^2 x}{\de t\;^2}\dfrac{\de y}{\de x}+(\dfrac{\de x}{\de t})^2\dfrac{\de^2 y}{\de x\;^2} \end{align}$