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I need to know wich answer is right

https://i.stack.imgur.com/QcxdH.jpg

I tried to solve it using recursivity but I didn't get any one of them

  • $y_1=\sqrt{x}$
  • $y'=\frac{1}{2\sqrt{x}}=\frac{1}{2y_1}$
  • $y_2=\sqrt{x+\sqrt{x}}=\sqrt{x+y_1}$ ;
  • $y_2'=\frac{1+y_1'}{2\sqrt{x+y_1}}=\frac{1+y_1'}{2y_2}=\frac{1+\frac{1}{2y_1}}{2y_2}=\frac{1+2y_1}{4y_1y_2}$
  • $y_3=\sqrt{x+\sqrt{x+\sqrt{x}}}=\sqrt{x+y_2}$
  • $y_3'=\frac{1+y_2'}{2y_3}=\frac{1+\frac{1+2y_1}{4y_1y_2}}{2y_3}=\frac{1+2y_1+4y_1y_2}{8y_1y_2y_3}$
  • .......
  • $y_n=\sqrt{x+\sqrt{x+\sqrt{x+...}}}=\sqrt{x+y_{n-1}}$
  • $y_n'=\frac{1+y_{n-1}'}{2y_n}=\frac{1+\sum_{i=1}^{n-1}2^i\prod_{j=1}^iy_j}{2^n\prod_{i=1}^{n}y_i}$
Darth Geek
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philomath213
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    notice that $y^2 = x+y$ - then use implicit differentiation – WW1 Dec 16 '16 at 03:41
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    It looks like you are taking a sequence of functions. What sort of convergence properties does it have? Also, the comment above mine is very nice. – The Count Dec 16 '16 at 03:41
  • the problem is mentioned in the image above – philomath213 Dec 16 '16 at 03:45
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    Related question talking about the convergence of $\sqrt{c+\sqrt{c+\sqrt{c+\dots}}}$. It is worth mentioning also that when $x<0$ your "function" is undefined for real numbers. You might be able to come up with some meaning using complex numbers still though if you wish to extend your domain. – JMoravitz Dec 16 '16 at 03:46
  • So, what is $\frac{dy}{dx}$? Is it P, Q, R or S? – Did Dec 16 '16 at 13:37
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    The OP's lack of reaction is disappointing. It happens in the present case that the answers received, the accepted one included, point at a wrong answer of the exercise the OP was asked to solve... but, as they say, one cannot force somebody to be saved against their own will. – Did Dec 17 '16 at 13:49

3 Answers3

3

$y^2=x+y$ now

$2ydy/dx=1+dy/dx$

bulbasaur
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As @WW1 mentioned in his comments, $$y=\sqrt {x+{\sqrt {x+\cdots}}} $$ $$\Rightarrow y=\sqrt {x+y} $$ $$\Rightarrow y^2=x+y $$ Now using the techniques of implicit differentiation, we have $$2y\frac{dy}{dx} = 1+\frac{dy}{dx} $$ Rearranging, we get, $$\Rightarrow \frac{dy}{dx} =\frac {1}{2y-1} $$ Hope it helps.

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$$y^2-y=x$$ then $$(2y-1)y'=1.$$