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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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
















-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











$endgroup$



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














-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











$endgroup$





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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 8:18









F.A.

9612719




9612719










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


















  • $begingroup$
    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










3 Answers
3






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









$endgroup$





















    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









    $endgroup$





















      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









      $endgroup$




















        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









        $endgroup$


















          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









          $endgroup$
















            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









            $endgroup$



            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












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 26 at 8:14









            JimmyK4542JimmyK4542

            41.3k245107




            41.3k245107























                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









                $endgroup$


















                  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









                  $endgroup$
















                    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









                    $endgroup$



                    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












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 26 at 8:17









                    Jimmy SabaterJimmy Sabater

                    3,054325




                    3,054325























                        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









                        $endgroup$


















                          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









                          $endgroup$
















                            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









                            $endgroup$



                            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












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 26 at 8:28









                            BadAtGeometryBadAtGeometry

                            188215




                            188215















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