Question Number 51446 by gunawan last updated on 27/Dec/18
$${f}\left({x}\right)=\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${f}^{'} \left(−\mathrm{2}\right)=... \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 27/Dec/18
![y^2 =((1−x^2 )/(1+x^2 )) 2y.(dy/dx)=(((1+x^2 )(−2x)−(1−x^2 )(2x))/((1+x^2 )^2 )) (dy/dx)=2(√((1−x^2 )/(1+x^2 )))[((−2x−2x^3 −2x+2x^3 )/((1+x^2 )^2 ))] (dy/dx)=2(√((1−x^2 )/(1+x^2 ))) [((−4x)/((1+x^2 )^2 ))] ((dy/dx))_(at x=−2) =2(√((1−4)/5)) ((8/(25)))=((2×(√3) i×8)/(25(√5)))=((16(√3) i)/(25(√5)))](Q51447.png)
$${y}^{\mathrm{2}} =\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{2}{y}.\frac{{dy}}{{dx}}=\frac{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(−\mathrm{2}{x}\right)−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{2}{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\left[\frac{−\mathrm{2}{x}−\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{2}{x}^{\mathrm{3}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right] \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }}\:\left[\frac{−\mathrm{4}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\right] \\ $$$$\left(\frac{{dy}}{{dx}}\right)_{{at}\:{x}=−\mathrm{2}} \:=\mathrm{2}\sqrt{\frac{\mathrm{1}−\mathrm{4}}{\mathrm{5}}}\:\left(\frac{\mathrm{8}}{\mathrm{25}}\right)=\frac{\mathrm{2}×\sqrt{\mathrm{3}}\:{i}×\mathrm{8}}{\mathrm{25}\sqrt{\mathrm{5}}}=\frac{\mathrm{16}\sqrt{\mathrm{3}}\:{i}}{\mathrm{25}\sqrt{\mathrm{5}}} \\ $$