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If y = tan-1[1+cosx1-cosx], find dddydx

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प्रश्न

If y = `tan^-1[sqrt((1 + cos x)/(1 - cos x))]`, find `("d"y)/("d"x)`

योग
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उत्तर

`tan^-1[sqrt((1 + cos x)/(1 - cos x))]`

Put `1 + cos x = 2 cos^2  x/2`

`1 - cos x = 2 sin^2  x/2`

∵ `cos x = 2 cos^2  x/2 - 1 = 1 - 2 sin^2  x/2`

= `tan^-1[sqrt((2cos^2 (x/2))/(2sin^2 (x/2)))]`

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

= `tan^-1 [cot(x/2)]`

= `tan^-1 [tan(pi/2- x/2)]      ...[cot theta = tan(pi/2 - theta)]`

= `pi/2 - x/2          ...[tan^-1 (tan theta) = theta]`

Differentiating w. r. t. x, we get

`("d"y)/("d"x) = "d"/("d"x) pi/2 -  "d"/("d"x) x/2`

∴ `("d"y)/("d"x) = 0 - 1/2 "d"/("d"x) x`

∴ `("d"y)/("d"x) = -1/2`

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अध्याय 2.1: Differentiation - Short Answers I

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