Question Number 61818 by Kunal12588 last updated on 09/Jun/19 | ||
$$\mathrm{Find}\:\:\:\frac{{dy}}{{dx}}\:\: \\ $$ $${y}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right),\:\mathrm{0}<{x}<\mathrm{1} \\ $$ | ||
Commented byPrithwish sen last updated on 09/Jun/19 | ||
$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\left[\frac{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\right]^{\mathrm{2}} }}\:\frac{−\mathrm{2x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$=\frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2x}}\:\frac{\left(−\mathrm{4x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$=\frac{−\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$ | ||
Commented byKunal12588 last updated on 09/Jun/19 | ||
$${see}\:{my}\:{way}... \\ $$ $${y}=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\:\:\:{let}\:{x}=\mathrm{tan}\:\alpha\Rightarrow\alpha=\:\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$ $${y}=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$ $$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha}}×\left(−\mathrm{sin}\:\mathrm{2}\alpha\right)×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{−\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$ | ||
Commented byPrithwish sen last updated on 09/Jun/19 | ||
$$\mathrm{Beautiful} \\ $$ | ||