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Differentiate sin-1(2cosx+3sinx13) w.r. to x - Mathematics and Statistics

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प्रश्न

Differentiate `sin^-1((2cosx + 3sinx)/sqrt(13))` w.r. to x

योग
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उत्तर

Let y = `sin^-1((2cosx + 3sinx)/sqrt(13))`

= `sin^-1((2cosx)/sqrt(13) + (3sinx)/sqrt(13))`

Put `2/sqrt(13)` = sin t and `3/sqrt(13)` = cos t

Also, sin2t + cos2t = `4/13 + 9/13` = 1

and tan t = `2/3`

∴ t = `tan^-1(2/3)`

∴ y = sin–1(sin t . cos x + cos t . sin x)

= sin–1[sin(t + x)]

= t + x

= `tan^-1(2/3) + x`

Differentiating w. r. t. x, we get

`("d"y)/("d"x) = "d"/("d"x)[tan^-1(2/3) + x]`

= 0 + 1

= 1

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अध्याय 2.1: Differentiation - Short Answers II

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