Question

$if \: y = \sqrt{x + \sqrt{x + \sqrt{x + .......... \infty \: } } } \\ show \: that \: (2y - 1) \: dy \div dx = 1$
PLZZ ANSWER IT FAST .​

Answer 1

Question :

If $y = \sqrt{x + \sqrt{x + \sqrt{x + ... \infty } } }$

Show that $(2y-1)\dfrac{dy}{dx}=1$

Formula :

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

• Differentiation Formula's

$1) \dfrac{d(sinx)}{dx} = cosx$

$2) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

$3) \dfrac{d(constant)}{dx} = 0$

Solution :

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

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

Now squaring on both sides

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

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

Now differentiate with respect to x

$\implies2y \dfrac{dy}{dx} - \dfrac{dy}{dx} = 1$

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

Hence proved !

Answer 2

Formula Required :

$\tt \red\star \: when \: y = {x}^{n} \\ \tt then \: \frac{dy}{dx} = {(nx)}^{n - 1}$

Solution :

we have given,

$\tt If\: y = \sqrt{x + \sqrt{x + \sqrt{x..... \infty } } }$

$\implies \tt y = \sqrt{x + \sqrt{x + \sqrt{x..... \infty } } }$

we can also write as,

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

Squaring both sides,we get.

$\implies \tt {y}^{2} = x + y.$

$\implies \tt {y}^{2} - y = x.$

Now, differentiate below equation with respect to x, we get.

$\tt \implies{( 2 \times y \times 1) }^{2 - 1} - {(x \times 1)}^{1 - 1} = 1.$

$\implies \tt 2y \frac{dy}{dx} - {x}^{0} \frac{dy}{dx} = 1$

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

Taking common dy / dx.

$\tt \implies \frac{dy}{dx} (2y - 1) = 1.$

Hance Proved.

$\rule{200}3$

Some Formula :

• y = a^x
• dy/dx = a^x log e^a.

• y = tanx
• dy/dx = sec²x.

• y= sec x
• dy/dx = sec x . tan x

• y= cot x
• dy/dx = -cosec ²x

• y= cosec x
• dy/dx = -cosec x . cot x.