Question

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

• A. $\frac{dy}{dx}=\frac{(1+y)\cos x+y\sin x}{1+3y+\cos x-\sin x}$
• B. $\frac{dy}{dx}=\frac{(1+y)\cos x+y\sin x}{1+y+\cos x-\sin x}$
• C. $\frac{dy}{dx}=\frac{(1+y)\cos x+y\sin x}{1+2y+\cos x-\sin x}$
• D. None of these

Answer 1

Answer: (C)

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

we can write the equation as $y=\frac{\sin x}{1+\frac{\cos x}{1+y}}$

$\Rightarrow$ $y=\frac{\sin x\times (1+y)}{1+y+\cos x}$

$\Rightarrow$ $y\times (1+y+\cos x)-\sin x\times (1+y)=0$

$\Rightarrow$ $y+y^2+y\cos x-y\sin x-\sin x=0$

Differentiating this equation with respect to x.

$\Rightarrow$ $\frac{dy}{dx}+2\times y\times \frac{dy}{dx}+\frac{dy}{dx}\times (\cos x-\sin x)-y\times (\sin x+\cos x)-\cos x=0$

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

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