Question

If
$y = \sqrt{sinx + \sqrt{sinx + \sqrt{sinx - - - \infty } } }$

find dy/dx

Please provide complete solution​

Answer 1

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

Given function is

$\rm :\longmapsto\:y = \sqrt{sinx + \sqrt{sinx + \sqrt{sinx + - - - \infty } } }$

On squaring both sides, we get

$\rm :\longmapsto\: {y}^{2} =sinx + \sqrt{sinx + \sqrt{sinx + \sqrt{sinx + - - - \infty } } }$

$\rm :\longmapsto\: {y}^{2} =sinx + y$

can be further rewritten as

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

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

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

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}sinx = cosx}}}$

and

$\purple{\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} }}}$

So, using these results, we get

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

$\rm :\longmapsto\:(2y - 1)\dfrac{dy}{dx}= cosx$

$\bf\implies \:\dfrac{dy}{dx} = \dfrac{cosx}{2y - 1}$

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SHORT CUT TRICK

$\rm :\longmapsto\:If \: y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + - - - \infty } } } \\ \ \\ \bf :\longmapsto\: then \: \dfrac{dy}{dx} \: = \: \dfrac{f'(x)}{2y - 1}$

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LEARN MORE

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

Answer 2

$\begin{gathered}\rm :\longmapsto\:If \: y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + - - - \infty } } } \\ \ \\ \bf :\longmapsto\: then \: \dfrac{dy}{dx} \: = \: \dfrac{f'(x)}{2y - 1} \end{gathered} </p><p>$