Question #d4611

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Answer 1 · 2017-10-08T01:09:32

$dy/dx=-2/(x^2+1)$ or $(-2x^(-1))/[x+x^(-1)]$

$y=cos^(-1) [(x-x^(-1))/(x+x^(-1))]$

= $cos^(-1) [(x^2-1)/(x^2+1)]$

Take cosine both sides,

$cosy=(x^2-1)/(x^2+1)$

Take differentiation both sides,

$-siny*dy=[2x*(x^2+1)-2x*(x^2-1)]*dx/(x^2+1)^2$

$-sqrt[1-(cosy)^2]*dy=(4x*dx)/(x^2+1)^2$

$-sqrt(1-[(x^2-1)/(x^2+1)]^2)*dy=(4x*dx)/(x^2+1)^2$

$-sqrt[(4x^2)/(x^2+1)^2]*dy=(4x*dx)/(x^2+1)^2$

$-(2x*dy)/(x^2+1)=(4x*dx)/(x^2+1)^2$

$dy/dx=(4x*dx)/(x^2+1)^2*-(x^2+1)/(2x)$

$dy/dx=-2/(x^2+1)$ or $(-2x^(-1))/[x+x^(-1)]$