If y={ x}{1+{ x}{1+{ x}{1+ x/}}}, then d y/d x is given by

Correct Answer is : ysin⁡x+(1+y)cos⁡x1+2y+cos⁡x−sin⁡xy x+(1+y) x/1+2 y+ x- x1+2y+cosx−sinxysinx+(1+y)cosx

Correct Answer is : ysin⁡x+(1+y)cos⁡x1+2y+cos⁡x−sin⁡xy x+(1+y) x/1+2 y+ x- x1+2y+cosx−sinxysinx+(1+y)cosx

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MHTCET 10 May 2024 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{\ldots}}}}

\frac{d y}{d x}

is given by

Solution:

Concept:

The infinite continued fraction repeats, so it can be written as a self-similar equation involving

y

Explanation:

Given

y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{\ldots}}}}

The tail after the first denominator is exactly

\frac{\cos x}{1+y}

Hence:

y=\frac{\sin x}{1+\frac{\cos x}{1+y}}

Simplify the denominator:

y=\frac{(1+y)\sin x}{1+y+\cos x}

Cross-multiplying:

y(1+y+\cos x)=(1+y)\sin x

y+y^2+y\cos x=\sin x+y\sin x

Differentiate both sides with respect to

x

y'+2yy'+y'\cos x-y\sin x=y'\sin x+(1+y)\cos x

Collect the

y'

terms:

y'(1+2y+\cos x-\sin x)=\cos x+y\cos x+y\sin x

Therefore:

\frac{dy}{dx}=\frac{y\sin x+(1+y)\cos x}{1+2y+\cos x-\sin x}

Answer:

Option A:

\frac{y\sin x+(1+y)\cos x}{1+2y+\cos x-\sin x}

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