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Differentiate tan-1(1-x2x) with respect tocos-1(2x1-x2), where x∈(12,1)

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प्रश्न

Differentiate `tan^-1 (sqrt(1 - x^2)/x)` with respect to`cos^-1(2xsqrt(1 - x^2))`, where `x ∈ (1/sqrt(2), 1)`

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उत्तर

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

We want to find `"du"/"dv" = (("du")/("dx"))/(("dv")/("dx"))`

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

Put x = `sintheta. (pi/2 < theta < pi/2)`

Then u = `tan^-1 (sqrt(1 - sin^2theta)/sintheta)`

= `tan^-1 (cot theta)`

= `tan^-1 {tan (pi/2 - theta)}`

= `pi/2 - theta`

= `pi/2 - sin^-1x`

Hence `"du"/"dx" = (-1)/sqrt(1 - x^2)`.

Now v = `cos^-1 (2x sqrt(1 - x^2))`

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

= `pi/2 - sin^-1 (2sintheta sqrt(1 - sin^2theta))`

= `pi/2 - sin^-1 (sin 2theta)`

= `pi/2 - sin^-1 {sin (pi - 2theta)}`  .......{Since  `pi/2` < 2θ < π]

= `pi/2 - (pi / 2theta)`

= `(-pi)/2 + 2theta`

⇒ v = `(-pi)/2 + 2sin^-1x`

⇒ `"dv"/"dv" = (("du")/("d"x))/(("dv")/("dx"))`

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

= `(-1)/2`

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अध्याय 5: Continuity And Differentiability - Solved Examples [पृष्ठ १०२]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 5 Continuity And Differentiability
Solved Examples | Q 23 | पृष्ठ १०२

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