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From Applied Differential Equations (first edition) by Spiegel,

Show that $$\frac{d^2y}{dx^2} = -\frac{d^2x}{dy^2}/\left(\frac{dx}{dy}\right)^3$$ is an identity. Hint: Differentiate both sides of $\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$ with resepct to $x$.

After looking at the hint, I got $$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dx}\left(\frac{dx}{dy}\right)$$ but I'm not sure how to proceed from here. Using $dx/dy = (dy/dx)^{-1}$ doesn't help, but otherwise I can't think of anyway to evaluate $\frac{d}{dx}\left(\frac{dx}{dy}\right)$.

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  • $\begingroup$ You should use the chain rule (as $dx/dy$ is a function of $y$). $\endgroup$ – Fabian Jan 1 '16 at 21:03
  • $\begingroup$ @Fabian: Ah, I think I was thinking that since $dx/dy$ was a function of $y$, differentiating it w.r.t. $x$ would be differentiating a constant, forgetting that $y$ is a function of $x$. $\endgroup$ – rwbogl Jan 1 '16 at 21:16
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From the hint, $$\frac{dy}{dx} = \left(\frac{dx}{dy}\right)^{-1}$$ Differentiate both sides with respect to $x$ to obtain $$\frac{d^2y}{dx^2} = -\left(\frac{dx}{dy}\right)^{-2}\frac{d}{dx}\left(\frac{dx}{dy}\right)$$ Thanks to Fabien's prodding, remember that $\frac{dx}{dy}$ is a function of $y$, and apply the chain rule and the hint again to get $$\frac{d}{dx}\left(\frac{dx}{dy}\right) = \frac{d^2x}{dy^2}\frac{dy}{dx} = \frac{d^2x}{dy^2}\left(\frac{dx}{dy}\right)^{-1}$$ Substituting this yields $$\frac{d^2y}{dx^2} = \frac{d^2x}{dy^2}/\left(\frac{dx}{dy}\right)^3$$

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With $f(g(y))=y$ you get in the first two derivatives $$ f'(g(y))g'(y)=1 $$ and $$ f''(g(y))g'(y)^2+f'(g(y))g''(y)=0 $$ Multiply with $g'(y)$ to obtain $$ f''(g(y))g'(y)^3=-g''(y) $$ and with $x=g(y)$, $y=f(x)$ $$ f''(x)=-g'(y)^{-3}g''(y) $$

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