# How do I calculate this second derivative? [closed]

Suppose $\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt}$. How do I calculate $\frac{d^2{y}}{dt^2}$?

• If this is homework, please say so. It will help you if you show your working too :) – Shaun Nov 13 '13 at 19:18

$\frac{d^2y}{dt^2} = \frac{d}{dt}\frac{dy}{dt} = \frac{d}{dt}\left[\frac{dy}{dx}\frac{dx}{dt}\right] = \frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2 + \frac{dy}{dx}\frac{d^2x}{dt^2}$,

where I have used the product rule in the brackets. You can think of $\frac{dx}{dt}$ as just some function of $t$ and $y=y(x(t),t)$, so $\frac{dy}{dx} = \frac{d}{dx}y(x(t),t)$. Looking at it this way may help justify the use of chain rule and subsequently product rule.

• Thanks @Jeremy. It helped a lot. – Sida Nov 15 '13 at 14:17

HINT: You can use the product rule.

• Why does it have to exist? – LASV Nov 13 '13 at 19:15

From the given equation

$\frac{dy}{dt} = \frac{dy}{dx}\frac{dx}{dt} \tag{1}$

we must infer that $y = y(x)$ is a function of $x$, and $x = x(t)$ is a function of $t$; it's really the most obvious way that (1) makes sense. This being the case, we have

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

As has been corroborated by Jeremy Upsal's answer.

Well, sometimes redundancy helps, so I hope this does. Cheerio,

and as always,

Fiat Lux!!!