If `"y"="tan"^(-1)"\ "[(sqrt(1+"x"^2-sqrt(1-"x"^2)))/(sqrt(1+"x"^2+)"\ "sqrt("\ "1-"x"^2)),]"\ f i n d\ \ "("dy")/("dx")`

Options

A. `(x^(2))/(sqrt(1-x^(4)))`
B. `(x^(2))/(sqrt(1+x^(4)))`
C. `(x)/(sqrt(1+x^(4)))`
D. `(x)/(sqrt(1-x^(4)))`

Answer

D

Explanation

Put `x^(2) = cos 2 theta` in the given equation, we get
`y = tan^(-1)""(sqrt(1+costheta)-sqrt(1-cos 2theta))/(sqrt(1+cos 2theta )+sqrt(1-cos 2theta))`
` = tan^(-1)""(cos theta-sin theta)/(cos theta + sin theta)`
` = tan^(-1)tan((pi)/(4)-theta)`
`rArr y = (pi)/(4)-theta = (pi)/(4) -(1)/(2) cos^(-1)x^(2)`
`rArr (dx)/(dx) = 0 -(1)/(2)(-((2x))/(sqrt(1-x^(4))))=(x)/(sqrt(1-x^(4)))`