If y = \(\sqrt{sinx + \sqrt{sinx + \sqrt{sinx + ……. to \infty}}}\), find \(dy\over dx\).

Solution :

Since by deleting a single term from an infinite series, it remains same. Therefore, the given function may be written as

y = \(\sqrt{sin x + y}\)

Squaring on both sides,

\(\implies\)  \(y^2\)  = sin x + y

By using differentiation of infinite series,

Differentiating both sides with respect to x,

2y \(dy\over dx\) =cosx +  \(dy\over dx\)

\(\implies\) \(dy\over dx\)\((2y – 1)\) = cos x

\(\implies\) \(dy\over dx\) = \(cos x\over {2y – 1}\)

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