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प्रश्न
Differentiate the following w.r.t. x : `sin^-1((4sinx + 5cosx)/sqrt(41))`
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उत्तर
Let y = `sin^-1((4sinx + 5cosx)/sqrt(41))`
= `sin^-1[(sinx)((4)/(sqrt(41))) + (cosx)((5)/(sqrt(41)))]`
Since, `(4/sqrt(41))^2 + (5/sqrt(41))^2 = (16)/(41) + (25)/(41) = 1`,
we can write, `(4)/sqrt(41) = cos ∞ and (5)/sqrt(41) = sin ∞`.
∴ y = sin–1 (sin x cos ∞ + cos x sin ∞)
= sin–1[sin(x + ∞)]
= x + ∞, where ∞ is a constant
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(x + ∞)`
= `"d"/"dx"(x) + "d"/"dx"(∞)`
= 1 + 0
= 1.
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