Question

Please solve the question

$y = \sqrt{tanx + \sqrt{tanx + \sqrt{tanx + - - - - \infty } } }$

Find dy/dx​

Answer 1

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

Given function is

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

So, it can be rewritten as

$\rm :\longmapsto\:y = \sqrt{tanx + y}$

On squaring both sides ,we get

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

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

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

$\rm :\longmapsto\:2y\dfrac{dy}{dx} = \dfrac{d}{dx}y + \dfrac{d}{dx}tanx$

$\rm :\longmapsto\:2y\dfrac{dy}{dx} = \dfrac{dy}{dx}+ {sec}^{2}x$

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

$\\ \rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \frac{ {sec}^{2}x}{2y - 1}}}$

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Short Cut Method

$\rm :\longmapsto\:y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + - - - \infty } } }$

then

$\\ \rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \frac{ f'(x)}{2y - 1}}}$

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More to Know

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

Answer 2

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

Given function is

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

So, it can be rewritten as

$\rm :\longmapsto\:y = \sqrt{tanx + y}$

On squaring both sides ,we get

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

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

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

$\rm :\longmapsto\:2y\dfrac{dy}{dx} = \dfrac{d}{dx}y + \dfrac{d}{dx}tanx$

$\rm :\longmapsto\:2y\dfrac{dy}{dx} = \dfrac{dy}{dx}+ {sec}^{2}x$

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

$\\ \rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \frac{ {sec}^{2}x}{2y - 1}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Short Cut Method

$\rm :\longmapsto\:y = \sqrt{f(x) + \sqrt{f(x) + \sqrt{f(x) + - - - \infty } } }$

then

$\\ \rm\implies \:\boxed{\tt{ \dfrac{dy}{dx} = \frac{ f'(x)}{2y - 1}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

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