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I know solving $\frac{d}{dy}(\frac{dx}{dy})=\frac{d}{dy}(\frac{1}{\frac{dy}{dx}}) = \frac{-\frac{d}{dy}(\frac{dy}{dx})}{(\frac{dy}{dx})^2} = - \frac{d^2y}{dx^2}(\frac{dy}{dx})^{-3}$

But I like to know why its wrong to write $\frac{d}{dy}(\frac{dx}{dy})=\frac{dy(\frac{dx}{dy})-dx\frac{dy}{dy}}{(dy)^2}$

Thanks.

Rajesh Marndi
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    You cannot simply treat $\frac{\mathrm dx}{\mathrm dy}$ as a fraction like that. In the given solution, $\frac{\mathrm dx}{\mathrm dy}\times\frac{\mathrm dy}{\mathrm dx}=1$ by the chain rule, hence $\frac{\mathrm dx}{\mathrm dy}=\frac1{\frac{\mathrm dy}{\mathrm dx}}$. See this related question with several answers that you might find illuminating. – user170231 Mar 12 '21 at 17:39
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    You can treat $\frac{\mathrm dx}{\mathrm dy}$ as fraction with appropriate definition. You can look discussing and many sources for it in my answer https://math.stackexchange.com/questions/3819116/rigorously-whats-happening-when-i-treat-fracdydx-as-a-fraction/3819142#3819142 – zkutch Mar 12 '21 at 22:40

2 Answers2

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I do not know why you would write like that. But my guess would be you are treating $\frac{dx}{dy}$ as a ratio and are trying to use the quotient rule.

$$(\frac{u}{v})'=\frac{vu'-uv'}{v^2}$$

Which is fine, Except... you will not get what you wrote. You will instead get

$$ \frac{dy(\frac{d}{dy}(dx))-dx(\frac{d}{dy}(dy))}{(dy)^2} $$

Chaitanya Chavali
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  • So would this get the same result?

    Also I don't know what $\frac{d}{dy}(dx)$ and $\frac{d}{dy}(dy)$ would mean and result.

     – Rajesh Marndi Mar 13 '21 at 04:38
  • @RajeshMarndi I don't know if it would give the same result. And I don't think anyone does. For the exact reason you stated. No one knows the value of dx as a function. Hence they don't know what d(dx)/dy means. – Chaitanya Chavali Mar 13 '21 at 06:14
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\begin{eqnarray} d \left(\frac{dx}{dy}\right) &=& \frac{d^2x dy - dx d^2y}{dy^2} \\ &=& \frac{d^2x dy/dx^3 - d^2y/dx^2}{dy^2/dx^3} \end{eqnarray} $d^2 x$ is the small change of $dx$. when $x$ is a "input variable", you can set $d^2x = 0$. But when you describe x as function of other variable like x = x(t) and consider t as an input, you have to revive this term. by setting $d^2x = 0$

\begin{eqnarray} \frac{d}{dy} \left(\frac{dx}{dy}\right) &=& - \frac{ d^2y/dx^2}{dy^3/dx^3} \end{eqnarray}

The above consideration about $d^2x$ is illustrated in the following example. The transformation shown below may seem correct but it omits the dependence on $d^2x$ and yield incorrect result.

$$ \frac{d^2y}{dt^2} = \frac{d^2y}{dx^2} \frac{dx^2}{dt^2} $$

The term $d^2x$ has to be revived in this way.

$$ d^2y = d (dy) = d(y' dx) = dy' dx + y' d^2x $$ Now using the relation $dy'/dx = y''$ \begin{eqnarray} \frac{d^2y}{dt^2} &=& y'' \frac{dx^2}{dt^2} + y' \frac{d^2x}{dt^2} \\ \end{eqnarray} This is a correct transformation. In this example, if you state $ y'' = \frac{d}{dx} \left(\frac{dy}{dx}\right) $ and use relation $ d \left(\frac{dy}{dx}\right) = \frac{d^2y dx - dy d^2x}{dx^2} $ Then $$ \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2} - \frac{dy}{dx} \frac{d^2x}{dx^2} $$ The ambiguity occurs because you can either say $$ y'' = \frac{d^2 y}{dx^2} $$ or $$ y'' = \frac{d^2 y}{dx^2} - \frac{dy}{dx} \frac{d^2x}{dx^2} $$ To avoid this ambiguity, we should define y'' to refer only to the case where $d^2 x$ is zero.

Hayashi Yoshiaki
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