Question

If $y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+.....\infty }}}}}$, then $y^{\prime }\left(0\right)$ is

• A. equal to $0$
• B. equal to $\frac{1}{2}$
• C. equal to $1$
• D. non existent

Answer 1

The correct option is B equal to $\frac{1}{2}$

$y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+.....\infty }}}}}$
here, $y\left(0\right)=0$
$\Rightarrow y=\frac{\sin x}{1+\frac{\cos x}{1+y}}$
$\Rightarrow y(1+y+\cos x)=(1+y)\sin x$
differentiating wrt x
$y^{\prime }(1+y+\cos x)+y(y^{\prime }-\sin x)=(1+y)\cos x+y^{\prime }\sin x$
$y^{\prime }\left(0\right)(1+y\left(0\right)+1)+y\left(0\right)\left(y^{\prime }\left(0\right)\right)=(1+y\left(0\right))$
$y^{\prime }\left(0\right)=\frac{1}{2}$