Differentiate `tan^(-1)((sqrt(1+x^2)-1)/x)` with respect to `sin^(-1)((2x)/(1+x^2)),`

Explanation

`u=tan^(-1)((sqrt(1+x^2)-1)/x)`

`x=tantheta`

`u=tan^(-1)[(sqrt(1+tan^2theta)-1)/tantheta]`

`=tan^(-1)[(sectheta-1)/tantheta]`

`=tan^(-1)[(1-costheta)/sintheta]`

`u=tan^(-1)[tantheta2]=theta/2`

`u=1/2tan^(-1)x`

`du/dx=d/dx[1/2tan^(-1)x]`

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