Question

Find `dy/dx` if `y = tan^-1 (sqrt((3 - x)/(3 + x)))`.

Answer 1

`y = tan^-1 (sqrt((3 - x)/(3 + x)))`

Put x = 3 cos2θ

∴ θ = `1/2 cos^-1(x/3)`   ...(1)

`y = tan^-1(sqrt((3 - 3cos2θ)/(3 + 3cos2θ)))`

= `tan^-1(sqrt((3(1 - cos2θ))/(3(1 + cos2θ))))`

= `tan^-1(sqrt((2sin^2θ)/(2cos^2θ)))`   ...[∵ 1 − cos2θ = 2sin²θ and 1 + cos2θ = 2cos²θ]

∴ = `tan^-1(sqrt(tan^2θ))`

∴ = tan⁻¹(tan²θ)

∴ y = θ   ...[∵ tan⁻¹ (tanθ) = θ]

∴ `y = 1/2 cos^-1 (x/3)`   ...[From (1)]

Differentiate w.r.t. x

`dy/dx = 1/2 d/dx [cos^-1(x/3)]`

= `1/2 xx -1/(sqrt (1 - (x/3)^2)) xx d/dx(x/3)`

= `1/2 xx (-1)/(sqrt(1 - x^2/9)) xx 1/3 xx 1`

= `-1/6 xx 3/sqrt(9 - x^2)`

= `-1/(2sqrt(9 - x^2))`