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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

  • 10
    That's a complicated way to write $\dfrac{\sqrt{1+4x}-1}{2}$ – 2017-01-17
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    Square both sides $y^2 = x+\sqrt{x+\cdots} =x+y$ and use implicit differentiation. – 2017-01-17

3 Answers 3

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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: $$ \frac{d}{dx}\,f(x) = \frac{1}{2\,f(x)-1}.$$

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    @egreg: I am assuming that the given problem is over $\mathbb{R}^+$, otherwise the definition of $f$ makes no sense. Over $\mathbb{R}^+$, the given function is differentiable. – 2017-01-17
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    Well, you should *prove* it is differentiable. – 2017-01-17
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    @egreg: that follows from usual inequalities and the concavity of $\sqrt{x}$, I do not think that is the interesting part, and I will leave it to the OP. – 2017-01-17
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$ y = \sqrt{x+\left(\sqrt{{x}+\sqrt{x+\cdots}}\right)}$

$y = \sqrt{x + y}$

Then $y^2 = x + y$

Now find derivative.

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

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

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

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

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    Haha, I too posted the same answer late only to look above. – 2017-01-17
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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 $y = \displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$