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