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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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Answer» `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))` |
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