`y=sin^(-1)""(2x)/(1+x^(2)),-1lexle1`

1.	`y=sin^(-1)""(2x)/(1+x^(2)),-1lexle1`
Answer» Correct Answer - `(2)/(1+x^(2))` `y=sin^(-1)((2x)/(1+x^(2)))` Let `x = tan theta.` Therefore, `y=sin^(-1)((2tan theta)/(1+tan^(2)theta))` `=sin^(-1)(sin 2 theta)` `=2theta` `=2tan^(-1)x` `therefore" "(dy)/(dx=(d)/(dx)(2 tan^(-1)x)` `=2(d)/(dx)(tan^(-1)x)` `(2)/(1+x^(2))`

Discussion

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