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Differentiate the following w.r.t. x : sin2[cot-1(1+x1-x)] - Mathematics and Statistics

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प्रश्न

Differentiate the following w.r.t. x : `sin^2[cot^-1(sqrt((1 + x)/(1 - x)))]`

योग
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उत्तर

Let y = `sin^2[cot^-1(sqrt((1 + x)/(1 - x)))]`

Put x = cosθ. Thenθ = cos–1x and
`sqrt((1 + x)/(1 - x)) = sqrt((1 + cosθ)/(1 - cosθ)`

= `sqrt((2cos^2(θ/2))/(2sin^2(θ/2)`

= `sqrt(cot^2(θ/2)`

= `cot(θ/2)`

∴ `cot^-1sqrt((1 + x)/(1 - x))`

= `cot^-1[cot(θ/2)]`

= `θ/(2)`

= `(1)/(2)cos^-1x`

∴ y = `sin^2(1/2 cos^-1x)`

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

= `2sin(1/2cos^-1x)."d"/"dx"sin(1/2cos^-1x)`

= `2sin(1/2cos^-1x).cos(1/2cos^-1x)."d"/"dx"(1/2cos^-1x)`

= `sin[2(1/2cos^-1x)] xx (1)/(2)."d"/"dx"(cos^-1x)`

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

= `sin(sin^-1sqrt(1 - x^2)) xx (-1)/(2sqrt(1 - x^2))   ...[∵ cos^-1x = sin^-1 sqrt(1 - x^2)]`

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

= `-(1)/(2)`.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 1: Differentiation - Miscellaneous Exercise 1 (II) [पृष्ठ ६४]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 4.2 | पृष्ठ ६४

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