If y=sqrt{sin (log_{2} x)+sqrt{sin (log_{2} x | vedclass

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(C) Given the equation $y=\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\ldots \infty}}$.Since the expression repeats infinitely,we can write it as

Vedclass · Accepted Answer

$\frac{\cos (\log_{2} x)}{x(2 y-1)}$

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(C) Given the equation $y=\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\ldots \infty}}$.Since the expression repeats infinitely,we can write it as $y=\sqrt{\sin (\log_{2} x)+y}$.Squaring both sides,we get $y^2 = \sin (\log_{2} x) + y$,which simplifies to $y^2 - y = \sin (\log_{2} x)$.Differentiating both sides with respect to $x$:$\frac{d}{dx}(y^2 - y) = \frac{d}{dx}(\sin (\log_{2} x))$.$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{d}{dx}(\log_{2} x)$.Using the change of base formula $\log_{2} x = \frac{\ln x}{\ln 2}$,we have $\frac{d}{dx}(\log_{2} x) = \frac{1}{x \ln 2}$.Thus,$(2y - 1) \frac{dy}{dx} = \cos (\log_{2} x) \cdot \frac{1}{x \ln 2}$.$\frac{dy}{dx} = \frac{\cos (\log_{2} x)}{x \ln 2 (2y - 1)}$.Note: If the original expression meant $\log(2x)$,the derivative is $\frac{\cos(\log(2x))}{x(2y-1)}$. Given the options,the intended expression is $\log(2x)$.

If $y=\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\sqrt{\sin (\log_{2} x)+\ldots \infty}}}$,then $\frac{d y}{d x}=$

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