Question Number 164747 by mathlove last updated on 21/Jan/22
$${faind}\:\:\frac{{dy}}{{dx}} \\ $$$${sin}^{−\mathrm{1}} \left({xy}\right)={csc}^{−\mathrm{1}} \left({x}−{y}\right) \\ $$
Answered by mahdipoor last updated on 21/Jan/22
$${sin}^{−\mathrm{1}} \left({xy}\right)={sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{x}−{y}}\right)\Rightarrow \\ $$$${xy}=\frac{\mathrm{1}}{{x}−{y}}\Rightarrow{y}+{x}\frac{{dy}}{{dx}}=\frac{\mathrm{1}−\frac{{dy}}{{dx}}}{\left({x}−{y}\right)^{\mathrm{2}} }\Rightarrow \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}−{y}\left({x}−{y}\right)^{\mathrm{2}} }{\mathrm{1}+{x}\left({x}−{y}\right)^{\mathrm{2}} } \\ $$