Question Number 14513 by tawa tawa last updated on 01/Jun/17
$$\mathrm{If}\:\:\mathrm{y}^{\mathrm{2}} \left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:=\:\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}\:−\:\mathrm{y}^{\mathrm{4}} }{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{4}} } \\ $$
Answered by mrW1 last updated on 01/Jun/17
$${y}^{\mathrm{2}} =\frac{\mathrm{2}−\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }−\mathrm{1}\:\:\:\:...\left({i}\right) \\ $$$$\mathrm{1}+{y}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\mathrm{1}−{y}^{\mathrm{2}} =\mathrm{2}−\frac{\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\left(\mathrm{1}+{y}^{\mathrm{2}} \right)\left(\mathrm{1}−{y}^{\mathrm{2}} \right)=\mathrm{1}−{y}^{\mathrm{4}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=−\frac{\mathrm{2}×\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${y}\frac{{dy}}{{dx}}=−\frac{\mathrm{2}{x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${y}^{\mathrm{2}} \left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{4}} }×\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)} \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }×\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$$$\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{1}−{y}^{\mathrm{4}} }{\mathrm{1}−{x}^{\mathrm{4}} } \\ $$
Commented by tawa tawa last updated on 01/Jun/17
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$
Answered by sandy_suhendra last updated on 01/Jun/17
![I try with the other way y^2 =((1−x^2 )/(1+x^2 )) y=[((1−x^2 )/(1+x^2 ))]^(1/2) let U=((1−x^2 )/(1+x^2 )) (dU/dx)=((−2x(1+x^2 )−2x(1−x^2 ))/((1+x^2 )^2 ))=((−4x)/((1+x^2 )^2 )) ((dU/dx))^2 =((16x^2 )/((1+x^2 )^4 )) and y=U^(1/2) (dy/dU)=(1/2)U^(− (1/2)) ((dy/dU))^2 =(1/4)U^(−1) =((1+x^2 )/(4(1−x^2 ))) ((dy/dx))^2 =((dy/dU))^2 ×((dU/dx))^2 =((1+x^2 )/(4(1−x^2 )))×((16x^2 )/((1+x^2 )^4 )) =(1/((1−x^2 )(1+x^2 )))×((4x^2 )/((1+x^2 )^2 )) = ((1−y^4 )/(1−x^4 )) ((4x^2 )/((1+x^2 )^2 ))=1−y^4 is from (ii) of mrW1′s answer](Q14528.png)
$$\mathrm{I}\:\mathrm{try}\:\mathrm{with}\:\mathrm{the}\:\mathrm{other}\:\mathrm{way} \\ $$$$\mathrm{y}^{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{y}=\left[\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\mathrm{let}\:\mathrm{U}=\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{dU}}{\mathrm{dx}}=\frac{−\mathrm{2x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{−\mathrm{4x}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:\:\:\: \\ $$$$\left(\frac{\mathrm{dU}}{\mathrm{dx}}\right)^{\mathrm{2}} =\frac{\mathrm{16x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\mathrm{and}\:\:\:\mathrm{y}=\mathrm{U}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dU}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{U}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\left(\frac{\mathrm{dy}}{\mathrm{dU}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\mathrm{U}^{−\mathrm{1}} =\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)} \\ $$$$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{dy}}{\mathrm{dU}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{dU}}{\mathrm{dx}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }{\mathrm{4}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}×\frac{\mathrm{16x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}×\frac{\mathrm{4x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}−\mathrm{y}^{\mathrm{4}} }{\mathrm{1}−\mathrm{x}^{\mathrm{4}} } \\ $$$$\frac{\mathrm{4x}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{1}−\mathrm{y}^{\mathrm{4}} \:\:\:\mathrm{is}\:\mathrm{from}\:\left(\mathrm{ii}\right)\:\mathrm{of}\:\mathrm{mrW1}'\mathrm{s}\:\mathrm{answer}\:\:\:\:\: \\ $$
Commented by tawa tawa last updated on 01/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$