Question

if y=
$\sqrt{x + \sqrt{x + \sqrt{x + .... \infty \: then \: dy \div dx = } } }$


​

Answer 1

GIVEN :–

$\\ { \bold{y = \sqrt{x + \sqrt{x + \sqrt{x + .................} } } }}$

TO FIND :–

$\\ { \bold{ \: \: \dfrac{dy}{dx} = ? }}$

SOLUTION :–

$\\ \implies{ \bold{y = \sqrt{x + \sqrt{x + \sqrt{x + .......} } } }}$

$\\ \implies{ \bold{y = \sqrt{x + \left[ \sqrt{x + \sqrt{x + .......} } \right]}}}$

• We should write this as –

$\\ \implies{ \bold{y = \sqrt{x + \left(y \right)}}}$

$\\ \implies{ \bold{y = \sqrt{x + y}}}$

• Square on both sides –

$\\ \implies{ \bold{(y) {}^{2} = {x + y}}}$

$\\ \implies{ \bold{y{}^{2} - y = {x }}}$

• Now Differentiate with respect to 'x' –

$\\ \implies{ \bold{2y \dfrac{dy}{dx} - \dfrac{dy}{dx} = \dfrac{dx}{dx} }}$

$\\ \implies{ \bold{2y \dfrac{dy}{dx} - \dfrac{dy}{dx} = 1 }}$

$\\ \implies{ \bold{(2y -1) \dfrac{dy}{dx} = 1 }}$

$\\ \implies \large { \boxed{ \bold{ \dfrac{dy}{dx} = \dfrac{1}{2y - 1} }}}$

$\\ \rule{220}{2}$

Used identity :–

$\\ (1) \: { \bold{ \dfrac{d( {x}^{n}) }{dx} = n {x}^{n - 1} }}$

$\\ (2) \: { \bold{ \dfrac{d{x}}{dx} = 1 }}$

$\\ \rule{220}{2}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star{\sf{ \: \: y = \sqrt{x + \sqrt{x + \sqrt{x + ..........to \infty } } } }}$

${\bf{\blue{\underline{To\:Find:}}}}$

$\star{\sf{ \: \: \frac{dy}{dx} = ?}}$

${\bf{\blue{\underline{Now:}}}}$

The given series can be written as:

$: \implies{\sf{ y \: = \sqrt{x + y} }}$

Squaring both side,

$: \implies{\sf{ {y}^{2} \: = (\sqrt{x + y}) ^{2} }}$

$: \implies{\sf{ {y}^{2} \: = {x + y} }}$

$: \implies{\sf{ {y}^{2} -y \: = {x } }}$

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Differentiate w.r.t.x,

$: \implies{\sf{ 2y \frac{dy}{dx} - \frac{dy}{dx} \: = 1}}$

Take dy/dx common,

$: \implies{\sf{ \frac{dy}{dx} (2y - 1) \: = 1}}$

$: \implies{\sf{ \frac{dy}{dx} = \frac{1}{2y - 1} }}$

$\star \boxed{\sf{ { \purple{ Hence \: \: \frac{dy}{dx} = \frac{1}{2y - 1} }}}}$

________________________________________

${\orange{ \boxed{{ \bf{ \star \: formula \: used}}}}}$

${\implies{{ \sf{ \star \: \frac{d}{dx}( {x}^{n) } = n {x}^{n - 1} }}}}$

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Note:-

• When we are to find dy/dx in case y is given as an infinite series,then a term is deleted from an infinite series,it remains unaltered.