math_differentiation_doubt_questionsDifferential equations 1. show that $\frac{{d}^{2}x}{d{y}^{2}}=-\left(\frac{dy}{dx}{\right)}^{-3}\phantom{\rule{0.22em}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}$ $\frac{{d}^{2}x}{d{y}^{2}}=\frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dy}\left(\frac{dy}{dx}{\right)}^{-1}=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\frac{d}{dy}\left(\frac{dy}{dx}\right)$$\begin{array}{l}\\ ⟹\frac{{d}^{2}x}{d{y}^{2}}=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\phantom{\rule{0.22em}{0ex}}\frac{dx}{dy}\frac{d}{dx}\left(\frac{dy}{dx}\right)\end{array}$ $⟹\frac{{d}^{2}x}{d{y}^{2}}=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\phantom{\rule{0.22em}{0ex}}\left(\frac{dy}{dx}{\right)}^{-1}\frac{d}{dx}\left(\frac{dy}{dx}\right)=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\phantom{\rule{0.22em}{0ex}}\left(\frac{dy}{dx}{\right)}^{-1}\frac{{d}^{2}y}{d{x}^{2}}\phantom{\rule{0.22em}{0ex}}$ $⟹\frac{{d}^{2}x}{d{y}^{2}}=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\phantom{\rule{0.22em}{0ex}}\left(\frac{dy}{dx}{\right)}^{-1}\frac{d}{dx}\left(\frac{dy}{dx}\right)=\left(-1\right)\left(\frac{dy}{dx}{\right)}^{-2}\phantom{\rule{0.22em}{0ex}}\left(\frac{dy}{dx}{\right)}^{-1}\frac{{d}^{2}y}{d{x}^{2}}$ $⟹\frac{{d}^{2}x}{d{y}^{2}}=-\left(\frac{dy}{dx}{\right)}^{-3}\phantom{\rule{0.22em}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}$