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Differentiate the following w.r.t. x : tan-1[1+x2+x1+x2-x] - Mathematics and Statistics

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Question

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

Sum
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Solution

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

Put x = tanθ. Thenθ = tan–1x

∴ `(sqrt(1 + x^2) + x)/(sqrt(1 + x^2 - x)) = sqrt(1 + tan^2θ + tanθ)/(sqrt(1 + tan^2θ) - tanθ)`

= `"secθ + tanθ"/"secθ - tanθ"`

= `((1/cosθ) + (sinθ/cosθ))/((1/cosθ) - (sinθ/cosθ))`

= `(1 + sinθ)/(1 - sinθ)`

= `(1 - cos(pi/2 + θ))/(1 + cos(pi/2 + θ)`

= `(2sin^2(pi/4 + θ/2))/(2cos^2(pi/4 + θ/2)`

= `tan^2(pi/4 + θ/2)`

∴ `sqrt((sqrt(1 + x^2) + x)/(sqrt(1 + x^2) - x)`

= `tan(pi/4 + θ/2)`

∴ y = `tan^-1[tan(pi/4 + θ/2)]`

= `pi/4 + θ/2`

= `pi/4 + 1/2tan^-1x`

∴ `"d"/"dx"(pi/4) + (1)/(2)"d"/"dx"(tan^-1x)`

= `0 + (1)/(2) xx (1)/(1 + x^2)`

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

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Chapter 1: Differentiation - Miscellaneous Exercise 1 (II) [Page 64]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 1 Differentiation
Miscellaneous Exercise 1 (II) | Q 4.6 | Page 64

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