If $y = sqrt{x + sqrt{x+sqrt{x+ldots}}}$ prove that $frac{dy}{dx}=frac{1}{2y-1}$ [duplicate]

-6

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This question already has an answer here:

• 
  What's the derivative of: $ sqrt{x+sqrt{{x}+sqrt{x+cdots}}}$?
  
  3 answers
  
  

If $$y = sqrt{x + sqrt{x+sqrt{x+ldots}}}$$

prove that $$frac{dy}{dx}=frac{1}{2y-1}$$

Please help me to solve this problem.

derivatives

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edited Jan 26 at 8:18

F.A.

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

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marked as duplicate by Robert Z, Did, Theo Bendit, Misha Lavrov, max_zorn Jan 26 at 22:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
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  Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
  $endgroup$
  – Math_QED
  Jan 26 at 9:28
  

  
  
  
  

add a comment | 

-6

$begingroup$

This question already has an answer here:

• 
  What's the derivative of: $ sqrt{x+sqrt{{x}+sqrt{x+cdots}}}$?
  
  3 answers
  
  

If $$y = sqrt{x + sqrt{x+sqrt{x+ldots}}}$$

prove that $$frac{dy}{dx}=frac{1}{2y-1}$$

Please help me to solve this problem.

derivatives

share|cite|improve this question

edited Jan 26 at 8:18

F.A.

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

$endgroup$

marked as duplicate by Robert Z, Did, Theo Bendit, Misha Lavrov, max_zorn Jan 26 at 22:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
  $endgroup$
  – Math_QED
  Jan 26 at 9:28
  

  
  
  
  

add a comment | 

-6

-6

-6

$begingroup$

This question already has an answer here:

• 
  What's the derivative of: $ sqrt{x+sqrt{{x}+sqrt{x+cdots}}}$?
  
  3 answers
  
  

If $$y = sqrt{x + sqrt{x+sqrt{x+ldots}}}$$

prove that $$frac{dy}{dx}=frac{1}{2y-1}$$

Please help me to solve this problem.

derivatives

share|cite|improve this question

edited Jan 26 at 8:18

F.A.

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

$endgroup$

This question already has an answer here:

• 
  What's the derivative of: $ sqrt{x+sqrt{{x}+sqrt{x+cdots}}}$?
  
  3 answers
  
  

If $$y = sqrt{x + sqrt{x+sqrt{x+ldots}}}$$

prove that $$frac{dy}{dx}=frac{1}{2y-1}$$

Please help me to solve this problem.

This question already has an answer here:

• 
  What's the derivative of: $ sqrt{x+sqrt{{x}+sqrt{x+cdots}}}$?
  
  3 answers
  
  

derivatives

derivatives

share|cite|improve this question

edited Jan 26 at 8:18

F.A.

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

share|cite|improve this question

edited Jan 26 at 8:18

F.A.

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

share|cite|improve this question

share|cite|improve this question

edited Jan 26 at 8:18

F.A.

9612719

edited Jan 26 at 8:18

F.A.

9612719

edited Jan 26 at 8:18

F.A.

9612719

9612719

asked Jan 26 at 8:10

Mark HenryMark Henry

44

asked Jan 26 at 8:10

Mark HenryMark Henry

44

asked Jan 26 at 8:10

Mark HenryMark Henry

44

44

marked as duplicate by Robert Z, Did, Theo Bendit, Misha Lavrov, max_zorn Jan 26 at 22:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Robert Z, Did, Theo Bendit, Misha Lavrov, max_zorn Jan 26 at 22:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
  $endgroup$
  – Math_QED
  Jan 26 at 9:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
  $endgroup$
  – Math_QED
  Jan 26 at 9:28
  

  
  
  
  

$begingroup$
Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
$endgroup$
– Math_QED
Jan 26 at 9:28

$begingroup$
Don't we need to prove the expression converges? Don't we need to prove that the function is differentiable?
$endgroup$
– Math_QED
Jan 26 at 9:28

add a comment | 

3 Answers
3

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oldest

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5

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Hint: Since $$y = sqrt{x+sqrt{x+sqrt{x+cdots}}},$$ we can write $$y = sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

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answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

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add a comment | 

1

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Since

$$ y = sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

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answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

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0

$begingroup$

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$boxed{y' = frac{1}{2y - 1}}$

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

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add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

5

$begingroup$

Hint: Since $$y = sqrt{x+sqrt{x+sqrt{x+cdots}}},$$ we can write $$y = sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

share|cite|improve this answer

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

$endgroup$

add a comment | 

5

$begingroup$

Hint: Since $$y = sqrt{x+sqrt{x+sqrt{x+cdots}}},$$ we can write $$y = sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

share|cite|improve this answer

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

$endgroup$

add a comment | 

5

5

5

$begingroup$

Hint: Since $$y = sqrt{x+sqrt{x+sqrt{x+cdots}}},$$ we can write $$y = sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

share|cite|improve this answer

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

$endgroup$

Hint: Since $$y = sqrt{x+sqrt{x+sqrt{x+cdots}}},$$ we can write $$y = sqrt{x+y},$$ i.e. $$y^2 = x+y.$$

Can you finish the problem from here?

share|cite|improve this answer

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

share|cite|improve this answer

share|cite|improve this answer

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

answered Jan 26 at 8:14

JimmyK4542JimmyK4542

41.3k245107

41.3k245107

add a comment | 

add a comment | 

1

$begingroup$

Since

$$ y = sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

share|cite|improve this answer

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

$endgroup$

add a comment | 

1

$begingroup$

Since

$$ y = sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

share|cite|improve this answer

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

$endgroup$

add a comment | 

1

1

1

$begingroup$

Since

$$ y = sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

share|cite|improve this answer

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

$endgroup$

Since

$$ y = sqrt{ x + y } $$, then

$$ y^2 = x + y $$

and using implicit differentiation with respect to x we obtain $2yy' = 1+y' $

share|cite|improve this answer

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

share|cite|improve this answer

share|cite|improve this answer

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

answered Jan 26 at 8:17

Jimmy SabaterJimmy Sabater

3,054325

3,054325

add a comment | 

add a comment | 

0

$begingroup$

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$boxed{y' = frac{1}{2y - 1}}$

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

$endgroup$

add a comment | 

0

$begingroup$

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$boxed{y' = frac{1}{2y - 1}}$

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

$endgroup$

add a comment | 

0

0

0

$begingroup$

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$boxed{y' = frac{1}{2y - 1}}$

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

$endgroup$

And to finish it off for you (based on JimmyK4542 and Jimmy Sabater's answers):

$2yy' = 1 + y'$

$2yy' - y' = 1$

$(2y - 1)y' = 1$

$boxed{y' = frac{1}{2y - 1}}$

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

share|cite|improve this answer

share|cite|improve this answer

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

answered Jan 26 at 8:28

BadAtGeometryBadAtGeometry

188215

188215

add a comment | 

add a comment | 

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