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Question
Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0
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Solution
Let y = `tan^-1 ((sqrt(1 + x^2) - 1)/x)` and z = tan–1x.
Put x = tan θ.
∴ y = `tan^-1 ((sqrt(1 + tan^2 theta) - 1)/tan theta)` and z = tan–1(tan θ) = θ
⇒ `tan ((sqrt(sec theta) - 1)/tan) = tan^-1 ((sec theta - 1)/tan theta)`
= `tan^-1 ((1/(cos theta) - 1)/((sin theta)/(cos theta))) = tan^-1 ((1 - cos theta)/sin theta)`
⇒ `tan^-1 ((2 sin^2 theta/2)/(2 sin theta /2 cos theta/2)) = tan^-1 ((sin theta/2)/(cos theta/2))`
⇒ y = `tan^-1 (tan theta/2)`
⇒ y = `theta/2`
Differentiating both parametric functions w.r.t. θ
`"dy"/("d"theta) = 1/2 * "d"/("d"theta) (theta)` and `"dz"/("d"theta) = "d"/("d"theta) (theta)`
= `1/2 * 1`
= `1/2` and `"dz"/("d"theta)` = 1
∴ `"dy"/"dz" = ("dy"/("d"theta))/("dz"/("d"theta))`
= `(1/2)/1`
= `1/2`.
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