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Question Number 61818 by Kunal12588 last updated on 09/Jun/19

Find   (dy/dx)    y = sin^(−1) (((1−x^2 )/(1+x^2 ))), 0<x<1

$$\mathrm{Find}\:\:\:\frac{{dy}}{{dx}}\:\: \\ $$ $${y}\:=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right),\:\mathrm{0}<{x}<\mathrm{1} \\ $$

Commented byPrithwish sen last updated on 09/Jun/19

(dy/dx) = (1/(√(1−[(((1−x^2 ))/((1+x^2 )))]^2 ))) ((−2x(1+x^2 )−2x(1−x^2 ))/((1+x^2 )^2 ))  =(((1+x^2 ))/(2x)) (((−4x))/((1+x^2 )^2 ))  =((−2)/((1+x^2 )))

$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\left[\frac{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\right]^{\mathrm{2}} }}\:\frac{−\mathrm{2x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$=\frac{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{2x}}\:\frac{\left(−\mathrm{4x}\right)}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$ $$=\frac{−\mathrm{2}}{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)} \\ $$

Commented byKunal12588 last updated on 09/Jun/19

see my way...  y= sin^(−1) (((1−x^2 )/(1+x^2 )))    let x=tan α⇒α= tan^(−1)  x  y= sin^(−1) (cos 2α)  ⇒(dy/dx)=(1/(√(1−cos^2  2α)))×(−sin 2α)×2×(1/(1+x^2 ))=((−2)/(1+x^2 ))

$${see}\:{my}\:{way}... \\ $$ $${y}=\:\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\:\:\:{let}\:{x}=\mathrm{tan}\:\alpha\Rightarrow\alpha=\:\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$ $${y}=\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$ $$\Rightarrow\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\alpha}}×\left(−\mathrm{sin}\:\mathrm{2}\alpha\right)×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\frac{−\mathrm{2}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$

Commented byPrithwish sen last updated on 09/Jun/19

Beautiful

$$\mathrm{Beautiful} \\ $$

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