Write a Limit to calculate $f'(0)$












1












$begingroup$


Let $f(x) = frac {2}{1+x^2} $



I need to write a limit to calculate $f'(0)$.



I think I have the basic understanding. Any help would be greatly appreciated.



d=delta and so far what I have is



$f'(x)$= lim (f(x+dx)-f(x))/dx

(dx)->0



((2/1+(x+dx)^2)-(2/1+x^2))/dx



((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx



((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx



((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx



(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)



(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)



(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)



that's as far as I have gotten. Any input would be great.






limits functions derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    f ' (o) derivative*
    $endgroup$
    – user637291
    Jan 22 at 19:24






  • 1




    $begingroup$
    Is this $$f(x)=frac{1}{1+x^2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:24






  • 1




    $begingroup$
    What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
    $endgroup$
    – John Coleman
    Jan 22 at 19:26






  • 1




    $begingroup$
    @Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:29






  • 1




    $begingroup$
    Yes we all miss that factor $2$, thank you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:31
















1












$begingroup$


Let $f(x) = frac {2}{1+x^2} $



I need to write a limit to calculate $f'(0)$.



I think I have the basic understanding. Any help would be greatly appreciated.



d=delta and so far what I have is



$f'(x)$= lim (f(x+dx)-f(x))/dx

(dx)->0



((2/1+(x+dx)^2)-(2/1+x^2))/dx



((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx



((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx



((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx



(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)



(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)



(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)



that's as far as I have gotten. Any input would be great.






limits functions derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    f ' (o) derivative*
    $endgroup$
    – user637291
    Jan 22 at 19:24






  • 1




    $begingroup$
    Is this $$f(x)=frac{1}{1+x^2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:24






  • 1




    $begingroup$
    What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
    $endgroup$
    – John Coleman
    Jan 22 at 19:26






  • 1




    $begingroup$
    @Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:29






  • 1




    $begingroup$
    Yes we all miss that factor $2$, thank you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:31














1












1








1


1



$begingroup$


Let $f(x) = frac {2}{1+x^2} $



I need to write a limit to calculate $f'(0)$.



I think I have the basic understanding. Any help would be greatly appreciated.



d=delta and so far what I have is



$f'(x)$= lim (f(x+dx)-f(x))/dx

(dx)->0



((2/1+(x+dx)^2)-(2/1+x^2))/dx



((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx



((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx



((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx



(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)



(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)



(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)



that's as far as I have gotten. Any input would be great.






limits functions derivatives







share|cite|improve this question











$endgroup$




Let $f(x) = frac {2}{1+x^2} $



I need to write a limit to calculate $f'(0)$.



I think I have the basic understanding. Any help would be greatly appreciated.



d=delta and so far what I have is



$f'(x)$= lim (f(x+dx)-f(x))/dx

(dx)->0



((2/1+(x+dx)^2)-(2/1+x^2))/dx



((2/1+x^2+2xdx+dx^2)-(2/1+x^2))/dx



((2(1+x^2)-2(1+x^2+2xdx+dx^2))/(1+x^2+2xdx+dx^2)(1+x^2))/dx



((-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2))/dx



(-4xdx-2dx^2)/(1+x^2+2xdx+dx^2)(1+x^2)(dx)



(-2dx(2xdx+dx)/(1+x^2+2xdx+x^2)(1+x^2)(dx)



(-2(2xdx+dx)/(1+x^2+2xdx+dx^2)(1+x^2)



that's as far as I have gotten. Any input would be great.






limits functions derivatives




limits functions derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 22 at 21:34







user637291

















asked Jan 22 at 19:23









user637291user637291

62




62












  • $begingroup$
    f ' (o) derivative*
    $endgroup$
    – user637291
    Jan 22 at 19:24






  • 1




    $begingroup$
    Is this $$f(x)=frac{1}{1+x^2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:24






  • 1




    $begingroup$
    What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
    $endgroup$
    – John Coleman
    Jan 22 at 19:26






  • 1




    $begingroup$
    @Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:29






  • 1




    $begingroup$
    Yes we all miss that factor $2$, thank you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:31


















  • $begingroup$
    f ' (o) derivative*
    $endgroup$
    – user637291
    Jan 22 at 19:24






  • 1




    $begingroup$
    Is this $$f(x)=frac{1}{1+x^2}$$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:24






  • 1




    $begingroup$
    What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
    $endgroup$
    – John Coleman
    Jan 22 at 19:26






  • 1




    $begingroup$
    @Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:29






  • 1




    $begingroup$
    Yes we all miss that factor $2$, thank you!
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 22 at 19:31
















$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24




$begingroup$
f ' (o) derivative*
$endgroup$
– user637291
Jan 22 at 19:24




1




1




$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24




$begingroup$
Is this $$f(x)=frac{1}{1+x^2}$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:24




1




1




$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26




$begingroup$
What happens when you use your basic understanding with this problem? What have you tried? Where are you stuck?
$endgroup$
– John Coleman
Jan 22 at 19:26




1




1




$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29




$begingroup$
@Dr.SonnhardGraubner I think you missed a factor of $2$ in the numerator. :)
$endgroup$
– The Jade Emperor
Jan 22 at 19:29




1




1




$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31




$begingroup$
Yes we all miss that factor $2$, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 22 at 19:31










2 Answers
2






active

oldest

votes


















1












$begingroup$

it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
right) left( {x}^{2}+1 right) }}
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:39



















0












$begingroup$

$f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
$= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$






share|cite|improve this answer











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    2 Answers
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    2 Answers
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    1












    $begingroup$

    it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
    Can you finish?
    Ok, another hint:
    The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
    right) left( {x}^{2}+1 right) }}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:39
















    1












    $begingroup$

    it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
    Can you finish?
    Ok, another hint:
    The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
    right) left( {x}^{2}+1 right) }}
    $$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:39














    1












    1








    1





    $begingroup$

    it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
    Can you finish?
    Ok, another hint:
    The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
    right) left( {x}^{2}+1 right) }}
    $$






    share|cite|improve this answer









    $endgroup$



    it is $$frac{f(x+h)-f(x)}{h}=frac{frac{2}{1+(x+h)^2}-frac{2}{1+x^2}}{h}$$
    Can you finish?
    Ok, another hint:
    The numerator is given by $$-2,{frac {h left( h+2,x right) }{ left( {h}^{2}+2,xh+{x}^{2}+1
    right) left( {x}^{2}+1 right) }}
    $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 22 at 19:34









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    77.5k42866




    77.5k42866












    • $begingroup$
      We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:39


















    • $begingroup$
      We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
      $endgroup$
      – The Jade Emperor
      Jan 22 at 19:39
















    $begingroup$
    We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:39




    $begingroup$
    We should probably stop here unless OP asks for more clarification. We still dont know how much understanding the OP has...
    $endgroup$
    – The Jade Emperor
    Jan 22 at 19:39











    0












    $begingroup$

    $f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
    $= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      $f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
      $= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        $f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
        $= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$






        share|cite|improve this answer











        $endgroup$



        $f'(0) = displaystyle lim_{h to 0} dfrac{f(h) - f(0)}{h} = lim_{h to 0} left (dfrac{1}{h} right ) left ( dfrac{2}{1 + h^2} - 2 right )$
        $= lim_{h to 0} left (dfrac{1}{h} right ) left ( -dfrac{2h^2}{1 + h^2} right ) = lim_{h to 0} -dfrac{2h}{1 + h^2} = 0. tag 1$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 19:39

























        answered Jan 22 at 19:30









        Robert LewisRobert Lewis

        47.9k23067




        47.9k23067






























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