Question

If $\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....+\infty }}}$then $(2\mathrm{y}-1)\frac{\mathrm{dy}}{\mathrm{dx}}$is equal to

• A. $\sin x$
• B. $\cos x$
• C. $-\sin x$
• D. $-\cos x$

Answer 1

The correct option is B

$\cos x$

Find the derivative:

Step1 : Simplify the function

$$\begin{gathered} y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....+\infty }}} \\ {y}^2=\sin x+\sqrt{\sin x+\sqrt{\mathrm{sinx}+\sqrt{\sin x+.....+\infty }}} \\ {y}^2=\sin x+y[y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+.....+\infty }}}] \end{gathered}$$

Step2 : Find the $\frac{dy}{dx}$

Differentiation with respect to $x$

$\begin{array}{rcl}2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}& =& \cos \mathrm{x}+\frac{\mathrm{dy}}{\mathrm{dx}}\\ (2\mathrm{y}-1)\frac{\mathrm{dy}}{\mathrm{dx}}& =& \cos \mathrm{x}\end{array}$

Hence, the correct option is B.