What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$?

Question

let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :

$\displaystyle \frac{dy}{dx}$ ?.

Note: I have used the definition of derivative of the square root function but i don't succed .

Thank you for any help

Answer 1

I am going to assume that the given problem is over $\mathbb{R}^+$, otherwise the definition of $f$ makes no sense. Over $\mathbb{R}^+$, the given function is differentiable by the concavity of $g(x)=\sqrt{x}$.

Such function fulfills $f(x)^2 = x+f(x)$, hence by termwise differentiation $$ 2\,f'(x)\,f(x) = 1 + f'(x) $$ and: $$ \frac{d}{dx}\,f(x) = \frac{1}{2\,f(x)-1}.$$

Answer 2

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

$y = \sqrt{x + y}$

Then $y^2 = x + y$

Now find derivative.

$2y\frac{dy}{dx} = 1 + \frac{dy}{dx}$

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

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

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

Answer 3

Let$$y = \displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$$ Its an infinite series so adding one term more will have not affect its value

$$y = \displaystyle \sqrt{x+\sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}}$$

From above equations $$y = \sqrt{x+y}$$ $$y^2 = x+y$$ Differentiating w.r.t. $x$, we get $$2y\frac{dy}{dx}=1+\frac{dy}{dx}$$ $$(2y-1)\frac{dy}{dx} = 1$$ $$\frac{dy}{dx}=\frac{1}{2y-1}$$ $$\frac{dy}{dx}=\frac{1}{2y-1}$$

where $y = \displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$

Answer 4

Clarification to solve the question
The question posed is calcule Derivative $\sqrt{x+\sqrt{x+\sqrt{x+....}}}$
we put $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}} $
Before calculating the derivative we simplify the expression $y$
$y^{2}=x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}=x+y$
$y^{2}-y-x=0$
We solve the equation $y^{2}-y-x=0$
The solution to this equation is $y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4x}}{2}$
After simplifying the relationship y we calculate $\frac{dy}{dx}$
we've got $\frac{dy}{dx}=\frac{1}{2}.\frac{4}{2\sqrt{1+4x}}=\frac{1}{\sqrt{1+4x}}$
In the general case, if $y=\sqrt{f(x)+\sqrt{f(x)+\sqrt{f(x)+...}}}$
$\Rightarrow y^{2}=f(x)+y$
$y^{2}-y-f(x)=0$
$y=\frac{-b+\sqrt{\bigtriangleup }}{2a}=\frac{1+\sqrt{1+4f(x)}}{2}$
$\frac{dy}{dx}=\frac{1}{2}\frac{4\acute{f}(x)}{2\sqrt{1+4f(x)}}=\frac{\acute{f}(x)}{\sqrt{1+4f(x)}}$