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Suppose I have obtained $\frac{d^2 y}{dt^2}=A$ and $\frac{dx}{dt}=B$.Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?How is it wrong if it is , because I am getting different answers for $y=t^3$ and $x=t^2$ from this process and if I had obtained $\frac{dy}{dx}$ and then took its derivative to find the second derivative .

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  • $\begingroup$ Can you explain how you arrived at the equation $\frac {d^{2}y} {dx^{2}}=\frac A {B^{2}}$?. $\endgroup$ Jul 18, 2019 at 8:00
  • $\begingroup$ That is the step which I think I'm getting wrong.What I did , correct/incorrect is $\frac{dx^2}{dt^2}=B^2$ $\endgroup$ Jul 18, 2019 at 8:03
  • $\begingroup$ No. You cannot treat a derivative $\frac {dy} {dx}$ as if it is a ratio of two numbers. You have to follows the rules of Calculus. Also, the symbol $\frac {dx^{2}} {dt^{2}}$ has no meaning. You can only talk about $\frac {d^{2}x} {dt^{2}}$ $\endgroup$ Jul 18, 2019 at 8:23

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Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?

No, you can't.

You could be tempted to do this because the notation resembles (normal) powers/exponents, but they're not. More specifically: $$\underbrace{\frac{\mbox{d}^2x}{\mbox{d}t^2}=\frac{\mbox{d}}{\mbox{d}t}\left(\frac{\mbox{d}x}{\mbox{d}t}\right)}_{\text{second derivative}} \ne \underbrace{\left(\frac{\mbox{d}x}{\mbox{d}t}\right)^2=\frac{\mbox{d}x}{\mbox{d}t}\frac{\mbox{d}x}{\mbox{d}t}}_{\text{square of the derivative}}$$


Suppose I have obtained $\frac{d^2 y}{dt^2}=A$ and $\frac{dx}{dt}=B$.Can I square B and divide A by it to obtain $\frac{d^2 y}{dx^2}=\frac{A}{B^2}$ ?

With $A$ and $B$ as above, you have: $$\frac{A}{B^2}=\frac{\frac{d^2y}{dt^2}}{\left(\frac{dx}{dt}\right)^2}$$ and it 'ends' there. What you hope (?), is not valid: $$\frac{\frac{d^2y}{dt^2}}{\left(\frac{dx}{dt}\right)^2} \color{red}{\ne} \frac{\frac{d^2y}{dt^2}}{\frac{dx^2}{dt^2}} \color{red}{"="} \frac{d^2y}{dx^2}$$

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  • $\begingroup$ But I don't want the $d^2 x$ term in numerator I need $dx^2$ can't I get it by $(dx)^2$ $\endgroup$ Jul 18, 2019 at 8:06
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    $\begingroup$ You should not see the $\mbox{d}^2x$ as a loose part, it is part of the notation $\frac{\mbox{d}^2x}{\mbox{d}t^2}$ for the second derivative of $x$ with respect to $t$. $\endgroup$
    – StackTD
    Jul 18, 2019 at 8:07
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Your problem lies in the squaring of $B$.

Remember what that notation for a derivative actually means. It's essentially a short form of the "real" definition of the derivative, that is $$\frac{df}{dx} = \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$$ Now you can see why the notation is that way, the $d$ symbols represent infinitesimal differences in $f$ and $x$, the numerator and denominator of the limit respectively. If you replace $B$ by the proper limit notation it will be obvious that squaring it (squaring the limit) is not the same as the double derivative, that is applying the derivative-limit method to the result of the derivative-limit method applied to $f$.

Edit: Here's a counter-example for you. Let $f = -\cos x$. $$\frac{d^2f}{dx^2} = \frac{d}{dx}\left[\frac{df}{dx}\right] = \frac{d}{dx}\left[\sin x\right] = \cos x \neq \sin^2 x = \left(\frac{df}{dx}\right)^2$$

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  • $\begingroup$ Squaring a limit is not same as double derivative, agreed but I'm not even trying to find double derivative of x with respect to the parameter , I have just found the first derivative and then squared it. $\endgroup$ Jul 18, 2019 at 8:20
  • $\begingroup$ Right but your formula doesn't work unless the square of the first derivative is the second derivative. How else would $A/B^2$ cancel to make $d^2y/dx^2$? Surely the only way would be if $B^2 = d^2x/dt^2$ which it is not. $\endgroup$ Jul 18, 2019 at 8:23
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The notation behind the second derivative is misleading (and, I would say, wrong). The meaning behind the notation is that $d^2y$ is supposed to be a double-differentiation, and $dx^2$ is the quantity $dx$ (the differential of x) squared. So, a more expansive way of writing it is this: $$\frac{d(d(y))}{(d(x))^2}$$ The problem with this is that it is actually wrong. To understand why, remember that the first derivative is a quotient: $\frac{dy}{dx}$. Therefore, taking the derivative of it will look like this: $$\frac{d\left(\frac{dy}{dx}\right)}{dx}$$ Therefore, for the second derivative, since there is a quotient in the thing you are taking the differential of, you need to use the quotient rule. Applying the quotient rule gives you the following: $$\frac{\frac{dx\, d(dy) - dy\, d(dx)}{dx^2}}{dx}$$ Substituting $d^2x$ for $d(dx)$ and $d^2y$ for $d(dy)$ gives you the following for an algebraically manipulable second derivative:

$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$ This is why your attempt does not work. Using differentials algebraically is fine, but you have to use a notation that makes sense of differentials. In your case, you would also need to know $\frac{d^2x}{dt^2}$ in order to make it all work (and the algebra is pretty ugly even then).

For more information about this, see the paper Extending the Algebraic Manipulability of Differentials.

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The equation can be solved using the algebraic double derivative.

$$ A=\frac{d\left[\frac{dy}{dt}\right]}{dt}=\frac{d^2y}{dt^2}-\frac{dy\ d^2t}{dt^3} $$

$$ B=\frac{dx}{dt} $$ And thus, similarly $$ B'=\frac{d^2x}{dt^2}-\frac{dx\ d^2t}{dt^3} $$

If you desire to find $y''$, the double derivative of $y$ with respect to $x$, first write the algebraic double derivative

$$ y''=\frac{d^2y}{dx^2}-\frac{dy\ d^2x}{dx^3} $$

One can manipulate the algebraic derivatives to get the following equation

$$ y''=\frac{A}{B^2}-B'B^3\frac{dy}{dx} $$

Which recovers the expected $A/B^2$ and an additional term.

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Before getting the first equation you must have found $C=\frac {dy} {dt}$. Use $\frac {dy} {dx}=(\frac {dy} {dt})/(\frac {dx} {dt})$ to find $\frac {dy} {dx}$. Let $z=\frac {dy} {dx}$. Use $\frac {dz} {dx} =(\frac {dz} {dt})/(\frac {dx} {dt})$ to find the second derivative.

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  • $\begingroup$ Yes sir , that surely gives the answer , but I'm not able to understand what's wrong in the above mentioned logic $\endgroup$ Jul 18, 2019 at 8:10

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