Question

If `y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))` , x²≤1, then find dy/dx.

Answer 1

`y=tan^(−1) ((sqrt(1+x^2)+sqrt(1−x^2))/(sqrt(1+x^2)−sqrt(1−x^2)))`

Putting x²=cos2θ, we have

`y=tan^(−1) ((sqrt(1+cos2θ)+sqrt(1−cos2θ))/(sqrt(1+cos2θ)−sqrt(1−cos2θ)))`

`y=tan^(−1) ((sqrt(2cos^2theta)+sqrt(2sin^2θ))/(sqrt(2cos^2θ)−sqrt(2sin^2θ)))`

`y=tan^(-1)((costheta+sintheta)/(costheta-sintheta))y`

`=tan^(-1)((1+tantheta)/(1-tantheta))` (Dividing the numerator and denominator by cosθ)

`y=tan^(-1)((tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta))`

`⇒y=tan^(−1)[tan(π/4+θ)]`

`⇒y=π/4+θ`

`∴ y=π/4+1/2cos^(−1)x^2              (x^2=cos2θ)`

Differentiating both sides with respect to x, we get

`dy/dx=0+1/2×(−1/sqrt(1−(x^2)^2))xx2x`

`⇒dy/dx=−x/sqrt(1−x^4)`