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DIfferentiate tan-1(1+x2-1x)w.r.t.tan-1(2x1-x21-2x2). - Mathematics and Statistics

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प्रश्न

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

बेरीज
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उत्तर

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

Then we want to find `"du"/"dv"`

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

Put x = tanθ

Then θ = tan–1 x

and

`(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)`

∴ u = `tan^-1[tan(θ/2)] = θ/(2) = (1)/(2)tan^-1x`

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

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

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

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

Put x = sin θ.

Then θ = sin–1x

and

`(2xsqrt(1 - x^2))/(1 - 2x^2)`

= `(2sinθsqrt(1 - sin^2θ))/(1 - 2sin^2θ)`

= `(2sinθcosθ)/(1 - 2sin^2θ)`

= `(sin2θ)/(cos2θ)`

= tan 2θ

∴ v = tan−1(tan2θ)

= 2θ

= 2sin−1x

∴ `"dv"/"dx" = 2"d"/"dx"(sin^-1x)`

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

∴ `"du"/"dx" = (("du"/"dx"))/(("dv"/"dx")`

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

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

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

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पाठ 1: Differentiation - Miscellaneous Exercise 1 (II) [पृष्ठ ६४]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 6.1 | पृष्ठ ६४

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