# Taking the second derivative of a parametric curve

I understand that for the parametric equations

$$\begin{align*}x&=f(t)\\ y&=g(t)\end{align*}$$

If $F(x)$ is the function with parameter removed then $\displaystyle F'(x) = \frac{\text{d}y}{\text{d}t}\big/\frac{\text{d}x}{\text{d}t}$

But the procedure for taking the second derivative is just described as " replace $y$ with dy/dx " to get

$$\frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)=\frac{\left[\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}t}\right)\right]}{\left(\frac{\text{d}x}{\text{d}t}\right)}$$

I don't understand the justification for this step. Not at all.

But that's all my book says on the matter then it launches in to plugging things in to this formula, and it seems to work well enough, but I don't know why.

I often find answers about question on differentials are beyond my level, I'd really like to get this, it'd mean a lot to me if someone could break it down.

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@nana: For semantic reasons it would be better if you used `\mathrm` instead of `\text` for setting the d's upright. – t.b. Jul 6 '11 at 2:35
@Theo: Thanks. duly noted...:) – Nana Jul 6 '11 at 2:45
@Nana: By the way, you could save yourself some typing work if you wrote `$\newcommand{\d}{\mathrm{d}}$` $\newcommand{\d}{\mathrm{d}}$ at the beginning of a post. Then writing `$\frac{\d}{\d t}$` gives $\frac{\d}{\d t}$. This would already pay if you had only three differential quotients – t.b. Jul 6 '11 at 2:51
@Theo:wow! didn't know that...Thanks again!! – Nana Jul 6 '11 at 2:58

## 2 Answers

Consider

$$\begin{align*} \frac{\text{d}^2y}{\text{d}x^2}&=\frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right)\\ &=\frac{\text{d}}{\text{d}t}\left(\frac{\text{d}y}{\text{d}x}\right).\frac{\text{d}t}{\text{d}x}\\ \end{align*}$$
where the last equality is as a result of applying the chain rule.

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Their justification is that you can use the same process for $\frac{dy}{dx}$ as for $Y$ since you can now consider $Y_2 = g_2(t) = \frac{dy}{dx}(t)$, that is, you once again have a parametric equation in terms of the parameter $t$, and the parametric equation for $x$ stays the same.

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