Question

Differentiate `tan^-1((sqrt(1 + x^2) - 1)/x)` w.r.t. `cos^-1(sqrt((1 + sqrt(1 + x^2))/(2sqrt(1 + x^2))))`

Answer 1

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

and v = `cos^-1(sqrt((1 + sqrt(1 + x^2))/(2sqrt(1 + x^2))))`

Then we want to find `(du)/(dv)`.

Put x = tan θ.

Then θ = tan^–1x.

Also, `(sqrt(1 + x^2) - 1)/x`

= `sqrt(1 + tan^2θ - 1)/tanθ`

= `(secθ - 1)/tanθ`

= `(1/cosθ - 1)/((sinθ/cosθ)`

= `( 1 - cosθ)/sinθ`

= `(2sin^2(θ/2))/(2sin(θ/2).cos(θ/2)`

= `tan(θ/2)`

and `(1 + sqrt(1 + x^2))/(2sqrt(1 + x^2)`

= `(1 + sqrt(1 + tan^2θ))/(2sqrt(1 + tan^2θ)`

= `(1 + secθ)/(2secθ)`

= `(1 + 1/cosθ)/((2/cosθ)`

= `(1 + cosθ)/(2)`

= `(2cos^2(θ/2))/(2)`

= `cos^2(θ/2)`

∴ `sqrt((1 + sqrt(1 + x^2))/(2sqrt(1 + x^2))) = cos(θ/2)`

∴ u = `tan^-1[tan(θ/2)]` and v = `cos^-1[cos(θ/2)]`

∴ u = `θ/2` and v = `θ/2`

∴ u = `1/2tan^-1x` and v = `1/2tan^-1x`

Differentiating u and v w.r.t. x, we get

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

= `1/2 xx 1/(1 + x^2)`

= `1/(2(1 + x^2)`

and `(dv)/dx = 1/2 d/dx (tan^-1x)`

= `1/2 xx 1/(1 + x^2)`

= `1/(2(1 + x^2)`

∴ `(du)/(dv) = ((du//dx))/((dv//dx)`

= `(1/(2(1 + x^2)))/(1/(2(1 + x^2))` = 1

Remark: u = `1/2tan^-1x` and v = `1/2tan^-1x`

∴ u = v

∴ `(du)/(dv) = d/(dv)(v)` = 1