If `x = log (1 + t^(2)) " and " y = t - tan^(-1)t, " then" dy/dx ` is equal to

Options

A. `e^(x) - 1`
B. `t^(2) - 1`
C. `(sqrt(e^(x) -1))/(2)`
D. ` e^(x) - y`

Answer

C

Explanation

Given , ` x = log (1 + t^(2)) " and " y = t - tan^(-1) t `
` (dx)/(dt) = (1)/(1 + t^(2)) 2t `
` (dy)/(dt) = 1 - (1)/(1 + t^(2)) = (t^(2))/(1 + t^(2))`
` therefore (dy)/(dx) = (t^(2) //1 + t^(2))/(2t//1 + t^(2)) = t//2 ` ...(i)
Also , ` x = log (1 + t^(2)) rArr t^(2) = r^(x) - 1` ...(iii)
From Eqs. (i) and (ii) , we get ` (dy)/(dx) = (sqrt(e^(x) -1))/(2) `