Question

Find the differentiation of infinite radicals,
$\dfrac{d}{dx} \left ( \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{... \infty } } } } \right)$
​

Answer 1

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

Given function is

$\rm :\longmapsto\: \sqrt{x + \sqrt{x + \sqrt{x + - - - \infty } } }$

Let assume that

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

On squaring both sides, we get

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

can be rewritten as

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

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

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

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

We know,

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

So, using this, we get

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

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

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

Short Cut Method

If

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

then

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

Additional Information :-

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

Answer 2

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:at \: first \: \: let \: \\ \\ \rm \: y = \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{ . \: . \: . \: \infty} } } } \: \: \: - - - (1) \\ \\ \rm \implies \: {y}^{2} = x + \sqrt{x + \sqrt{x + \sqrt{. \: . \: . \: \infty} } } \\ \\ \rm \implies \: {y}^{2} = x + \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{... \infty} } } } \\ \\ \rm \implies \: {y}^{2} = x + y \: \: \: \{ \sf \: by \: eqn.(1) \} \\ \\ \sf \: differentiate \: w.r.t \: \: \rm \: x \\ \\ \rm \: \frac{d}{dx} {y}^{2} = \frac{d}{dx}(x + y) \\ \\ \rm \implies \: 2y. \frac{dy}{dx} = \frac{d}{dx} x + \frac{d}{dx} \: y \\ \\ \rm \implies \: 2y. \frac{dy}{dx} = 1 + \frac{dy}{dx } \\ \\ \rm \implies \: 2y. \frac{dy}{dx} - \frac{dy}{dx} = 1 \\ \\ \rm \implies \: (2y - 1). \frac{dy}{dx} = 1 \\ \\ \rm \implies \: \frac{dy}{dx} = \frac{1}{2y - 1} \\ \\ \\ \orange{ \boxed{\therefore \rm \: \frac{d}{dx} \left ( \sqrt{x + \sqrt{x + \sqrt{x + \sqrt{... \infty} } } } \: \right ) = \frac{1}{2y - 1} }}\end{array}}}}$