If y-sin x-sin x-sin x- then d y-d x is equal toA- cos x-2 y-1B- 2 x-1cos x cos x C- cos n-2 x-1D- 2 y-1-cos x

The correct option is A cos x/2y 1y = √(sin x + y)⇒ y^2 y = sin x ∴ (2y 1) dy/dx = cos x ...

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Implicit Differentiation

If y=√sin x+√...

Question

If $y = \sqrt{s i n x + \sqrt{s i n x + \sqrt{s i n x + . . . . \infty}}}$ ,then $\frac{d y}{d x}$ is equal to

\frac{2 y - 1}{c o s x}

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\frac{c o s x}{2 y - 1}

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\frac{2 x - 1}{c o s x}

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\frac{c o s x}{2 x - 1}

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Solution

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Similar questions

I f y = \sqrt{s i n x + \sqrt{s i n x + \sqrt{s i n x + . . . . . \infty}}}, t h e n \frac{d y}{d x} i s e q u a l t o

y = \sqrt{sin x - \sqrt{sin x - \sqrt{sin x - . . . . . \infty}}}

\frac{d y}{d x} =

I f y = \sqrt{s i n x + \sqrt{s i n x + \sqrt{s i n x + . . . . . \infty}}}, t h e n \frac{d y}{d x} i s e q u a l t o

y = \sqrt{cos x + \sqrt{cos x + \sqrt{cos x + . . . . . . . \infty}}}

\frac{d y}{d x} = \frac{- sin x}{(2 y - 1)}

y = \sqrt{sin x + \sqrt{sin x + \sqrt{sin x + . . . . . . . . . . . \infty}}}

\frac{d y}{d x} = \frac{cos x}{2 y - 1}

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