Question

$\large \blue{ \underline{\maltese \textsf{ \:Find the derivative of : - }}}$
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$\boxed{ \sf \large \sqrt{\cos x \sqrt{ \cos \:x \sqrt{ \cos \: x \sqrt{ \cos \: x } .}. } .} \infty }$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given that

${\sqrt{\cos x \sqrt{ \cos \:x \sqrt{ \cos \: x \sqrt{ \cos \: x } .}. } .} \infty }$

Let assume that

$\rm :\longmapsto\:y = {\sqrt{\cos x \sqrt{ \cos \:x \sqrt{ \cos \: x \sqrt{ \cos \: x } .}. } .} \infty }$

On squaring both sides, we get

$\rm :\longmapsto\: {y}^{2} = {{cos x \sqrt{ \cos \:x \sqrt{ \cos \: x \sqrt{ \cos \: x } .}. } .} \infty }$

can be rewritten as

$\rm :\longmapsto\: {y}^{2} = {{cos x \times y} }$

$\rm :\longmapsto\: {y}^{2} - ycosx = 0$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}({y}^{2} - ycosx )= 0$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}(u - v) = \dfrac{d}{dx}u - \dfrac{d}{dx}v}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{d}{dx} {y}^{2} - \dfrac{d}{dx}y \: cosx = 0$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}$

and

$\boxed{ \bf{ \: \dfrac{d}{dx}(u. v) =u \dfrac{d}{dx}v + v \dfrac{d}{dx}u}}$

So, using these two results, we get

$\rm :\longmapsto\:2y\dfrac{d}{dx}y - \bigg(y\dfrac{d}{dx}cosx + cosx\dfrac{d}{dx}y\bigg) = 0$

We know,

$\boxed{ \bf{ \: \dfrac{d}{dx}cosx = - \: sinx}}$

So, using these results, we get

$\rm :\longmapsto\:2y\dfrac{dy}{dx} - \bigg( - ysinx + cosx\dfrac{dy}{dx} \bigg) = 0$

$\rm :\longmapsto\:2y\dfrac{dy}{dx} + ysinx - cosx\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:\dfrac{dy}{dx}(2y - cosx)= - ysinx$

$\rm :\implies\:\dfrac{dy}{dx} = \dfrac{ - y \: sinx}{2y - cosx}$

$\bf :\implies\:\dfrac{dy}{dx} = \dfrac{ y \: sinx}{cosx - 2y}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$