Finding $\dfrac{d^nx}{dy^n}$

Question

If $y$ is a function of $x$, then what is the relation between $\dfrac{d^nx}{dy^n}$ and $\dfrac{d^ny}{dx^n}$? If we were to talk about $\dfrac{dy}{dx}$ and $\dfrac{dx}{dy}$, then they both are reciprocals of each other.

We cannot simply take the reciprocal of $\dfrac{d^ny}{dx^n}$, can we?

And then still with $y=x^2$, see what happens for $n=3,4,5,\ldots$ — Théophile, Jul 14 '14 at 13:22
@O.L. $\dfrac{\pm 1}{8y^{3/2}}\dfrac{d^2y}{dx^2}=\dfrac{d^2x}{dy^2}$ — user1001001, Jul 14 '14 at 13:23
@Théophile for $n=2,3,4...$ we can say that $\dfrac{d^ny}{dx^n}.\dfrac{d^nx}{dy^n}=0$ as $\dfrac{d^ny}{dx^n}=0$... — user1001001, Jul 14 '14 at 13:28

Ben Grossmann · Accepted Answer · 2014-07-14T18:26:16.817

5

When in doubt, use the chain rule. Repeatedly differentiate both sides with respect to $y$ as follows: $$ y = f(x)\\ 1 = f'(x) \cdot \frac{dx}{dy} \implies \frac{dx}{dy} = 1/f'(x) \\ 0 = f''(x) \left(\frac{dx}{dy}\right)^2 + f'(x) \frac{d^2x}{dy^2} \implies\\ \frac{d^2x}{dy^2} = -\frac{f''(x)}{f'(x)} \left(\frac{dx}{dy}\right)^2 \implies\\ \frac{d^2x}{dy^2} = -\frac{f''(x)}{[f'(x)]^3} $$ That is, at a given point $(x,y)$, we have $$ \frac{d^2x}{dy^2} = -\frac{d^2y/dx^2}{(dy/dx)^3} $$

edited Jul 14 '14 at 18:26

answered Jul 14 '14 at 13:27

Ben Grossmann

225,327

Will we be able to generalize it for any n? – user1001001 Jul 14 '14 at 13:31
@user157130 I'm not sure how exactly this would generalize – Ben Grossmann Jul 14 '14 at 13:32
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@user157130 that being said, it is possible to continue this process indefinitely by differentiating both sides with respect to $y$ yet again. – Ben Grossmann Jul 14 '14 at 13:47
Yes..that is correct..and thanks for your help! – user1001001 Jul 14 '14 at 13:48

Finding $\dfrac{d^nx}{dy^n}$

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