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Let $y=y(x)$ be determined by the equation \begin{align*}\begin{cases} x=t-\sin{t}\\ y=1-\cos{t}.\end{cases} \end{align*} I understand the solution: $$\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dt}\frac{1}{\frac{dx}{dt}}=-\frac{1}{(1-\cos{t})^2}$$ But what's wrong with the following calculations: $$\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\frac{dt^2}{dx^2}=\frac{\cos{t}dt^2}{dt^2}\frac{dt^2}{((1-\cos{t})dt)^2}=\frac{\cos{t}}{(1-\cos{t})^2}$$

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  • $\begingroup$ your formula for $\frac{d^2y}{dx^2}$ like a chain's rule is wrong $\endgroup$ – janmarqz Dec 18 '14 at 4:59
  • $\begingroup$ A related question $\endgroup$ – Omnomnomnom Dec 19 '14 at 13:47
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By the use of the chain's rule you get $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$ then $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}.\qquad(1)$$

Now $\frac{d}{dt}\!\!\left(\frac{dy}{dt}\right)=\frac{d}{dt}\!\!\left(\frac{dy}{dx}\frac{dx}{dt}\right)=\frac{d}{dx}\left(\frac{dy}{dx}\frac{dx}{dt}\right)\frac{dx}{dt}=\frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{dy}{dx}\frac{d}{dx}\left(\frac{dx}{dt}\right)\frac{dx}{dt}$.

Then $$\frac{d^2y}{dt^2}= \frac{d^2y}{dx^2}\left(\frac{dx}{dt}\right)^2+\frac{dy}{dx}\frac{d^2x}{dt^2}.\qquad(2)$$ So upon substitution of $(1)$ in $(2)$ and solving for $\frac{d^2y}{dx^2}$ we get $$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{dt^2}\frac{dx}{dt}-\frac{d^2x}{dt^2}\frac{dy}{dt}}{\left(\frac{dx}{dt}\right)^3},$$ which can be used to solve your problem.

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  • $\begingroup$ Can I interpret $d^2y$ as $d(dy)=d(d(1-\cos{t}))$ and $dx^2=(dx)^2=(d(t-\sin{t}))^2$? $\endgroup$ – longtemps Dec 18 '14 at 6:37
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    $\begingroup$ @longtemps: No you should not. $\frac{d^2y}{dx^2}$ is the second derivative of $y$ with respect to $x.$ It is not the second derivative of $y$ with respect to $x,$ divided by the square of the derivative of $x$ with respect to $t.$ Interpreting things that way is what got you the wrong answer from before! $\endgroup$ – Cameron Buie Dec 18 '14 at 12:31
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Good question! :) The thing to remember is that differentiation is an operator. It isn't a scalar variable that can be tossed around. Here is an example:

As you know, the definition of the first derivative would be: $$\frac{dy}{dx} \approx \frac{\Delta( y)}{\Delta x}$$ However, we don't know what the change in $y$ is for a given change in $x$ directly. This is why we need to use a common variable, $t$. So the equation becomes: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \approx\frac{\frac{\Delta y}{\Delta t}}{\frac{\Delta x}{\Delta t}}$$ If we choose exactly the same bounds in $t$ for $x$ and $y$, this convention is sort of a crude way to describe it. What you are actually doing is this: $$\frac{dy}{dx}=\frac{\frac{d}{dt}(y)}{\frac{d}{dt}(x)}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ When we deal with the second derivative, the approximation would be: $$\frac{d^2y}{dx^2}\approx\frac{\Delta\left(\frac{\Delta y}{\Delta x}\right)}{\Delta x}$$ What we are approximating is the change in slope over the same interval $\Delta x$. Going back to the differentials, we get: $$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt} (\frac{dy}{dx})}{\frac{d}{dt}(x)}$$ In all these cases, the results of the differentiation are in terms of $t$, not $x$ or $y$, which is part of the issue.

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  • $\begingroup$ are ours answers equivalent? : ) $\endgroup$ – janmarqz Dec 2 '15 at 16:50

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