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If \[y=\dfrac{\sin x}{1+\dfrac{\cos }{1+\dfrac{\sin x}{1+\dfrac{\cos x}{1+\sin x........}}}}\], then find \[\dfrac{dy}{dx}\].

Last updated date: 23rd Jul 2024
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Hint: Simplify the denominator which is an infinite function and differentiate by applying product rule.

Here \[y\] is an infinite function. By observation, we can see the pattern that \[\sin x\] and \[\cos x\]appear alternatingly and hence, we can write \[y\] as
\[y=\dfrac{\sin x}{1+\dfrac{\cos x}{1+y}}\]
\[\Rightarrow y=\dfrac{\sin x}{\dfrac{1+y+\cos x}{1+y}}\]
\[\Rightarrow y=\dfrac{\sin x(1+y)}{1+y+\cos x}\]
\[\Rightarrow y(1+y+\cos x)=\sin x(1+y)\]
Differentiating both sides with respect to\[x\], we get,
\[(1+y+\cos x).\dfrac{dy}{dx}+y(1+\dfrac{dy}{dx}-\sin x)=(1+y)\cos x+\sin x(\dfrac{dy}{dx})\]
\[\Rightarrow (1+y+\cos x).\dfrac{dy}{dx}+y+y\dfrac{dy}{dx}-y\sin x=(1+y)\cos x+\sin x(\dfrac{dy}{dx})\]
Taking all the terms with \[\dfrac{dy}{dx}\]to one side and other terms to the other side, we get,
\[\Rightarrow (1+y+\cos x).\dfrac{dy}{dx}+y\dfrac{dy}{dx}-\sin x(\dfrac{dy}{dx})=(1+y)\cos x+y\sin x-y\]
\[\Rightarrow (1+y+\cos x-\sin x+y).\dfrac{dy}{dx}=(1+y)\cos x+y\sin x-y\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{y(cosx+\sin x-1)+\cos x}{1+2y+\cos x-\sin x}\]

Note: Generally students make a mistake of writing \[\dfrac{\sin x}{1+\dfrac{\cos }{1+\dfrac{sinx}{1+\cos x.....}}}\] as \[\dfrac{\sin x}{\dfrac{1+\cos x}{y}}\]which is wrong as the repetition starts after the addition of \[1\] in the denominator i.e. \[\dfrac{\sin x}{\dfrac{1+\cos x}{1+y}}\].