Question

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

Answer 1

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. $\dfrac{\sin x}{\dfrac{1+\cos x}{1+y}}$.