मराठी

Differentiate tan-1(1+x2-1x) w.r.t. tan–1x, when x ≠ 0

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प्रश्न

Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0

बेरीज
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उत्तर

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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पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ १११]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 53 | पृष्ठ १११

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