Question

Differentiate `tan^-1[(sqrt(1 + x^2) - 1)/x]` w.r. to `tan^-1[(2x sqrt(1 - x^2))/(1 - 2x^2)]`

Answer 1

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

Put x = tan θ, then θ = tan^–1x

∴ u = `tan^-1[(sqrt(1 + tan^2theta) - 1)/(tantheta)]`

= `tan^-1[(sqrt(sec^2theta) - 1)/(tantheta)]`

= `tan^-1[(sectheta - 1)/(tantheta)]`

= `tan^-1[(1/costheta - 1)/((sintheta)/(costheta))]`

= `tan^-1[(1 - costheta)/(sintheta)]`

= `tan^-1[(2sin^2(theta/2))/(2sin(theta/2)cos(theta/2))]`

= `tan^-1[tan  theta/2]`

= `theta/2`

∴ u = `1/2 tan^-1x`

Differentiating w.r.t. x, we get

`("d"u)/("d"x) = 1/((2(1 + x^2)`

Let v = `tan^-1[(2xsqrt(1 - x^2))/(1 - 2x^2)]`

Put x = sin θ, then θ = sin^–1x

∴ v = `tan^-1[(2sinthetasqrt(1 - sin^2theta))/(1 - 2sin^2theta)]`

= `tan^-1[(2sinthetacostheta)/(cos2theta)]`

= `tan^-1[(sin2theta)/(cos2theta)]`

= tan⁻¹ (tan 2θ) = 2θ

∴ v = 2sin⁻¹x

Differentiating w.r.t. x, we get

`("d"v)/("d"x) = 2/sqrt(1 - x^2)`

∴ `("d"u)/("d"x) = ((("d"u)/("d"x)))/((("d"v)/("d"x))`

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

= `sqrt(1 - x^2)/(4(1 + x^2))`