Question

If $y=\frac{\sin x}{1+\frac{\cos x}{1+\frac{\sin x}{1+\cos x...\infty }}}$ then $\frac{dy}{dx}=?$

• A. $\frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x}$
• B. $\frac{(1+y)\sin x+y\cos x}{1+2y+\cos x-\sin x}$
• C. $\frac{(1+y)\cos x-y\sin x}{1-2y-\cos x+sinx}$
• D. $\frac{(1+7)\cos x+y\sin x}{1+2y-cosx-\sin x}$

Answer 1

The correct option is A $\frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x}$

$y=\frac{\sin x}{1+\frac{\cos x}{\sin x}}$, then $\frac{dy}{dx}=?$

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

$y=\frac{\sin x}{\frac{(1+y)+\cos x}{(1+y)}}$ or $y=\frac{\left(\sin x\right)(1+y)}{(y+\cos x)}$

$\left(y\right)(y+\cos x)=\left(\sin x\right)(1+y)$

$y+y^2(\cos x-\sin x)=\sin x$

Differentiating wrt $x$

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

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

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