यदि `y=tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))]x^(2)le1`, तो `(dy)/(dx)` का ज्ञात कीजिये ।
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`y=tan^(-1)[(sqrt(1+x^(2))+sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))]`
`x^(2)=cos 2 theta` रखने पर,
`y = tan ^(-1)[(sqrt(1+cos 2 theta )+sqrt(1-cos 2 theta))/(sqrt(1+cos 2 theta)-sqrt(1-cos 2 theta))]`
`=tan^(-1)[(sqrt(2 cos 2 theta)+sqrt(2 sin 2 theta))/(sqrt(2 cos^(2)theta)-sqrt(2 sin ^(2)theta))]`
`=tan^(-1)[(cos theta+sin theta)/(cos theta – sin theta)]`
`=tan^(4)[(1+tan theta)/(1-tan theta)]`
`=tan^(-1)[(tan.(pi)/(4)+tan theta)/(1-tan .(pi)/(4)tan theta)]`
`=tan^(-1){tan ((pi)/(4)+theta)}`
`=(pi)/(4) + theta = (pi)/(4) + (1)/(2) cos^(-1)x^(2)`
`rArr (dy)/(dx)=-(1)/(2) (1)/(sqrt(1-(x^(2))^(2)))(2x)`
`=-(x)/(sqrt(1-x^(4))`