How we simplify equation with derivation












0












$begingroup$


Our teacher write :



$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$



We can write equation $(1)$ like this



$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$



but I can not understand how we find the equation $(2).$






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please write your equations using MathJax. Here's a tutorial.
    $endgroup$
    – md2perpe
    Nov 30 '18 at 17:53
















0












$begingroup$


Our teacher write :



$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$



We can write equation $(1)$ like this



$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$



but I can not understand how we find the equation $(2).$






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Please write your equations using MathJax. Here's a tutorial.
    $endgroup$
    – md2perpe
    Nov 30 '18 at 17:53














0












0








0





$begingroup$


Our teacher write :



$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$



We can write equation $(1)$ like this



$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$



but I can not understand how we find the equation $(2).$






calculus derivatives







share|cite|improve this question











$endgroup$




Our teacher write :



$(zz')' +2xz' = 0,quad 0 < x < infty tag {1}$



We can write equation $(1)$ like this



$(z^2)'' + 4xz' =0,quad 0 < x < infty tag{2}$



but I can not understand how we find the equation $(2).$






calculus derivatives




calculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 22:16









user376343

3,0532825




3,0532825










asked Nov 30 '18 at 17:46









N111N111

262




262












  • $begingroup$
    Please write your equations using MathJax. Here's a tutorial.
    $endgroup$
    – md2perpe
    Nov 30 '18 at 17:53


















  • $begingroup$
    Please write your equations using MathJax. Here's a tutorial.
    $endgroup$
    – md2perpe
    Nov 30 '18 at 17:53
















$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53




$begingroup$
Please write your equations using MathJax. Here's a tutorial.
$endgroup$
– md2perpe
Nov 30 '18 at 17:53










2 Answers
2






active

oldest

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0












$begingroup$

The key observation here is that
$$
(z^2)'' = 2(zz')'
$$
which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.



This means, if we multiply the first equation through by $2$, then we get
$$
2(zz')' + 4xz' = (z^2)' + 4xz'
$$
as expected.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Application of the Chain Rule :



    $$(z^2)’ = 2zz’$$



    Thus:



    $$(z^2)’ + 4xz = 0$$



    becomes



    $$2(zz’)’ + 4xz =0$$



    Divide through by 2 to get your first equation.






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      0












      $begingroup$

      The key observation here is that
      $$
      (z^2)'' = 2(zz')'
      $$
      which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.



      This means, if we multiply the first equation through by $2$, then we get
      $$
      2(zz')' + 4xz' = (z^2)' + 4xz'
      $$
      as expected.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        The key observation here is that
        $$
        (z^2)'' = 2(zz')'
        $$
        which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.



        This means, if we multiply the first equation through by $2$, then we get
        $$
        2(zz')' + 4xz' = (z^2)' + 4xz'
        $$
        as expected.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          The key observation here is that
          $$
          (z^2)'' = 2(zz')'
          $$
          which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.



          This means, if we multiply the first equation through by $2$, then we get
          $$
          2(zz')' + 4xz' = (z^2)' + 4xz'
          $$
          as expected.






          share|cite|improve this answer









          $endgroup$



          The key observation here is that
          $$
          (z^2)'' = 2(zz')'
          $$
          which can be derived using either the product rule or the chain rule. Using the chain rule, if we let $f(x) = x^2$, then we're looking to find $(f(z))' = f'(z)z'$, and since $f'(x) = 2x$, this is just $2zz'$. To get the second derivative, just apply the derivative again (but don't simplify): $(f(z))'' = (f'(z)z')' = (2zz')' = 2(zz')'$, where the last equality is true since the derivative is a linear operator.



          This means, if we multiply the first equation through by $2$, then we get
          $$
          2(zz')' + 4xz' = (z^2)' + 4xz'
          $$
          as expected.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 17:57









          user3002473user3002473

          3,38252239




          3,38252239























              0












              $begingroup$

              Application of the Chain Rule :



              $$(z^2)’ = 2zz’$$



              Thus:



              $$(z^2)’ + 4xz = 0$$



              becomes



              $$2(zz’)’ + 4xz =0$$



              Divide through by 2 to get your first equation.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Application of the Chain Rule :



                $$(z^2)’ = 2zz’$$



                Thus:



                $$(z^2)’ + 4xz = 0$$



                becomes



                $$2(zz’)’ + 4xz =0$$



                Divide through by 2 to get your first equation.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Application of the Chain Rule :



                  $$(z^2)’ = 2zz’$$



                  Thus:



                  $$(z^2)’ + 4xz = 0$$



                  becomes



                  $$2(zz’)’ + 4xz =0$$



                  Divide through by 2 to get your first equation.






                  share|cite|improve this answer









                  $endgroup$



                  Application of the Chain Rule :



                  $$(z^2)’ = 2zz’$$



                  Thus:



                  $$(z^2)’ + 4xz = 0$$



                  becomes



                  $$2(zz’)’ + 4xz =0$$



                  Divide through by 2 to get your first equation.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 17:59









                  ip6ip6

                  54839




                  54839






























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