STD 12 - 5. continuity and differentiation — JEE कक्षा 11 साइन्स — Question

$ = \frac{d}{{dx}}\left( {\frac{1}{{\frac{{dy}}{{dx}}}}} \right) \cdot \frac{{dx}}{{dy}} =  - \frac{1}{{{{\left( {\frac{{dy}}{{dx}}} \right)}^2}}} \cdot \frac{{{d^2}y}}{{d{x^2}}} \cdot \frac{1}{{\left( {\frac{{dy}}{{dx}}} \right)}}$

$=-\frac{1}{\left(\frac{d y}{d x}\right)^{3}} \cdot \frac{d^{2} y}{d x^{2}}=-\left(\frac{d y}{d x}\right)^{-3}\left(\frac{d^{2} y}{d x^{2}}\right) $