Answer
Verified
91.5k+ views
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. \[\dfrac{\sin x}{\dfrac{1+\cos x}{1+y}}\].
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. \[\dfrac{\sin x}{\dfrac{1+\cos x}{1+y}}\].
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
If the central portion of a convex lens is wrapped class 12 physics JEE_Main