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If $\mathrm{y}=\mathrm{f}(\mathrm{x}), \mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}$ and $\mathrm{q}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$, then what is $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}$ equal
to?
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to?
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Verified Answer
The correct answer is:
$-\frac{\mathrm{q}}{\mathrm{p}^{3}}$
As given, $y=f(x), p=\frac{d y}{d x}$ and $q=\frac{d^{2} y}{d x^{2}}$
$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{\mathrm{p}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=\frac{-1}{\mathrm{p}^{2}} \cdot \frac{\mathrm{dp}}{\mathrm{dy}}$
$\frac{\mathrm{dp}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{q}$
$\frac{d p}{d y}=\frac{d p}{d x} \cdot \frac{d x}{d y}=q \cdot \frac{1}{p}=\frac{q}{p}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \mathrm{y}^{2}}=\Rightarrow-\frac{1}{\mathrm{p}^{2}} \times \frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{q}}{\mathrm{p}^{3}}$
$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{\mathrm{p}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=\frac{-1}{\mathrm{p}^{2}} \cdot \frac{\mathrm{dp}}{\mathrm{dy}}$
$\frac{\mathrm{dp}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{q}$
$\frac{d p}{d y}=\frac{d p}{d x} \cdot \frac{d x}{d y}=q \cdot \frac{1}{p}=\frac{q}{p}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \mathrm{y}^{2}}=\Rightarrow-\frac{1}{\mathrm{p}^{2}} \times \frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{q}}{\mathrm{p}^{3}}$
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