Question Number 44623 by mondodotto@gmail.com last updated on 02/Oct/18
$$\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{{x}}+\boldsymbol{\mathrm{sin}}^{−\mathrm{1}} \boldsymbol{{y}}=\boldsymbol{\mathrm{c}} \\ $$$$\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}\:\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}+\sqrt{\frac{\mathrm{1}−\boldsymbol{{y}}^{\mathrm{2}} }{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }}=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
$$\frac{{d}}{{dx}}\left({sin}^{−\mathrm{1}} {x}+{sin}^{−\mathrm{1}} {y}\right)=\frac{{dc}}{{dx}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\frac{\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+\frac{{dy}}{{dx}}=\mathrm{0} \\ $$