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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?

Frank Gallagher
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  • What is $y$...? – IAmNoOne Mar 13 '15 at 01:14
  • y is a function of x, and x is in turn a function of t – Bamboo Mar 13 '15 at 01:16
  • $y$ is the solutions to a differential equation. Solving this problem is the first step to solving the differential equation. – Frank Gallagher Mar 13 '15 at 01:16
  • http://math.stackexchange.com/questions/432955/how-do-we-take-second-order-of-total-differential This is nearly the same as what you are asking – Bamboo Mar 13 '15 at 01:23
  • I don't completely understand the answers presented on that page, but I'm fairly certain that partial differentials should not be in the answer to this problem. – Frank Gallagher Mar 13 '15 at 01:32
  • The partial differentials only come in because the example is slightly more general as it involves a function in more than one varible. The general idea is the same though and you should be able to pick ou the answer you need if you dig enough. See the answer I provide for a more targeted response. – Bamboo Mar 13 '15 at 01:45

2 Answers2

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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:

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

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

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

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

Graham Kemp
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