Derivative of function and simplification The Next CEO of Stack OverflowFind the second...

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Derivative of function and simplification



The Next CEO of Stack OverflowFind the second derivative of $e^{x^{3}}+7x$Second Derivative of basic fraction using quotient ruleSimplifying Second DerivativesI need help finding the critical values of this function.How do I find the derivative of $y = frac{-5}{cos sqrt{{3x+2x^3}}}$?Derivatives of trigonometric functions: $y= frac{x sin(x)}{1+cos(x)}$Derivative of function $ y=[e^{x^2}-cot(ln(sqrt x+frac 1x))]^{sec(frac1x)} $Need clarification on a simplification step in a differentiation problemdifferentiate equations using multiple rulesHelp with a simplification












1












$begingroup$


I'm attempting to find the derivative of the following function:



$$frac{e^x - e^{-x}}{e^x + e^{-x}}$$



where I would use the quotient rule to find its derivative. In the end, I would obtain:



$$frac{(e^x + e^{-x})^2 -(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$



However, I checked with the calculator online and I found out that the equation I current have above can be further simplified to:



$$text{ Rewrite / Simplify }$$
$$=1-frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
$$text{ Simplify }$$
$$frac{4e^{2x}}{(e^{2x} +1)^2}$$



I'm having issues understanding the simplification step above.



My biggest question is, in the second image above, why couldn't I have $(e^x + e^{-x})^2$ cancel out $(e^x + e^{-x})^2$ at the denominator, which would leave me with just $-(e^x - e^{-x})^2$? Could anyone explain the simplification step to me as I am very much confused.




P.S. pardon me but I couldn't understand the syntax on writing equations here so I would opt to use screenshot instead.






calculus derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    "Derivative of a function", not "derivative of an equation".
    $endgroup$
    – Jean Marie
    Mar 16 at 6:10
















1












$begingroup$


I'm attempting to find the derivative of the following function:



$$frac{e^x - e^{-x}}{e^x + e^{-x}}$$



where I would use the quotient rule to find its derivative. In the end, I would obtain:



$$frac{(e^x + e^{-x})^2 -(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$



However, I checked with the calculator online and I found out that the equation I current have above can be further simplified to:



$$text{ Rewrite / Simplify }$$
$$=1-frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
$$text{ Simplify }$$
$$frac{4e^{2x}}{(e^{2x} +1)^2}$$



I'm having issues understanding the simplification step above.



My biggest question is, in the second image above, why couldn't I have $(e^x + e^{-x})^2$ cancel out $(e^x + e^{-x})^2$ at the denominator, which would leave me with just $-(e^x - e^{-x})^2$? Could anyone explain the simplification step to me as I am very much confused.




P.S. pardon me but I couldn't understand the syntax on writing equations here so I would opt to use screenshot instead.






calculus derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    "Derivative of a function", not "derivative of an equation".
    $endgroup$
    – Jean Marie
    Mar 16 at 6:10














1












1








1





$begingroup$


I'm attempting to find the derivative of the following function:



$$frac{e^x - e^{-x}}{e^x + e^{-x}}$$



where I would use the quotient rule to find its derivative. In the end, I would obtain:



$$frac{(e^x + e^{-x})^2 -(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$



However, I checked with the calculator online and I found out that the equation I current have above can be further simplified to:



$$text{ Rewrite / Simplify }$$
$$=1-frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
$$text{ Simplify }$$
$$frac{4e^{2x}}{(e^{2x} +1)^2}$$



I'm having issues understanding the simplification step above.



My biggest question is, in the second image above, why couldn't I have $(e^x + e^{-x})^2$ cancel out $(e^x + e^{-x})^2$ at the denominator, which would leave me with just $-(e^x - e^{-x})^2$? Could anyone explain the simplification step to me as I am very much confused.




P.S. pardon me but I couldn't understand the syntax on writing equations here so I would opt to use screenshot instead.






calculus derivatives







share|cite|improve this question











$endgroup$




I'm attempting to find the derivative of the following function:



$$frac{e^x - e^{-x}}{e^x + e^{-x}}$$



where I would use the quotient rule to find its derivative. In the end, I would obtain:



$$frac{(e^x + e^{-x})^2 -(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$



However, I checked with the calculator online and I found out that the equation I current have above can be further simplified to:



$$text{ Rewrite / Simplify }$$
$$=1-frac{(e^x - e^{-x})^2}{(e^x + e^{-x})^2}$$
$$text{ Simplify }$$
$$frac{4e^{2x}}{(e^{2x} +1)^2}$$



I'm having issues understanding the simplification step above.



My biggest question is, in the second image above, why couldn't I have $(e^x + e^{-x})^2$ cancel out $(e^x + e^{-x})^2$ at the denominator, which would leave me with just $-(e^x - e^{-x})^2$? Could anyone explain the simplification step to me as I am very much confused.




P.S. pardon me but I couldn't understand the syntax on writing equations here so I would opt to use screenshot instead.






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 Mar 17 at 3:33









rash

553115




553115










asked Mar 16 at 5:48









ElectricElectric

116127




116127








  • 2




    $begingroup$
    "Derivative of a function", not "derivative of an equation".
    $endgroup$
    – Jean Marie
    Mar 16 at 6:10














  • 2




    $begingroup$
    "Derivative of a function", not "derivative of an equation".
    $endgroup$
    – Jean Marie
    Mar 16 at 6:10








2




2




$begingroup$
"Derivative of a function", not "derivative of an equation".
$endgroup$
– Jean Marie
Mar 16 at 6:10




$begingroup$
"Derivative of a function", not "derivative of an equation".
$endgroup$
– Jean Marie
Mar 16 at 6:10










3 Answers
3






active

oldest

votes


















2












$begingroup$

I think your issue is that the computation in the last "cartridge" should be decomposed into $2$ steps :



First step :



The first expression can be written, taking a common denominator :



$$dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{4}{(e^x+e^{-x})^2}$$



Second step : Factor $e^{-x}$ in the denominator :



$$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$



We have obtained a $(e^{2x}-1)^2$ as a factor in the denominator.



Getting rid of $e^{-2x}$ in the denominator by transforming it into $e^{2x}$ in the numerator, explains the final result with $4e^{2x}$ in the numerator.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    thank you, it was a very nice explanation. I understand it now.
    $endgroup$
    – Electric
    Mar 16 at 6:31



















0












$begingroup$

Hint: Use that $$a^2-b^2=(a-b)(a+b)$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    So it's not even about derivative, as your question is that why:



    $$frac{a - b}{a} ne -b, a>0$$



    Well, they are just not equal, because in general $a-b+ab ne 0$






    share|cite|improve this answer









    $endgroup$














      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      I think your issue is that the computation in the last "cartridge" should be decomposed into $2$ steps :



      First step :



      The first expression can be written, taking a common denominator :



      $$dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{4}{(e^x+e^{-x})^2}$$



      Second step : Factor $e^{-x}$ in the denominator :



      $$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$



      We have obtained a $(e^{2x}-1)^2$ as a factor in the denominator.



      Getting rid of $e^{-2x}$ in the denominator by transforming it into $e^{2x}$ in the numerator, explains the final result with $4e^{2x}$ in the numerator.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        thank you, it was a very nice explanation. I understand it now.
        $endgroup$
        – Electric
        Mar 16 at 6:31
















      2












      $begingroup$

      I think your issue is that the computation in the last "cartridge" should be decomposed into $2$ steps :



      First step :



      The first expression can be written, taking a common denominator :



      $$dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{4}{(e^x+e^{-x})^2}$$



      Second step : Factor $e^{-x}$ in the denominator :



      $$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$



      We have obtained a $(e^{2x}-1)^2$ as a factor in the denominator.



      Getting rid of $e^{-2x}$ in the denominator by transforming it into $e^{2x}$ in the numerator, explains the final result with $4e^{2x}$ in the numerator.






      share|cite|improve this answer











      $endgroup$









      • 1




        $begingroup$
        thank you, it was a very nice explanation. I understand it now.
        $endgroup$
        – Electric
        Mar 16 at 6:31














      2












      2








      2





      $begingroup$

      I think your issue is that the computation in the last "cartridge" should be decomposed into $2$ steps :



      First step :



      The first expression can be written, taking a common denominator :



      $$dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{4}{(e^x+e^{-x})^2}$$



      Second step : Factor $e^{-x}$ in the denominator :



      $$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$



      We have obtained a $(e^{2x}-1)^2$ as a factor in the denominator.



      Getting rid of $e^{-2x}$ in the denominator by transforming it into $e^{2x}$ in the numerator, explains the final result with $4e^{2x}$ in the numerator.






      share|cite|improve this answer











      $endgroup$



      I think your issue is that the computation in the last "cartridge" should be decomposed into $2$ steps :



      First step :



      The first expression can be written, taking a common denominator :



      $$dfrac{(e^x+e^{-x})^2}{(e^x+e^{-x})^2}+dfrac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{(e^x+e^{-x})^2+(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=dfrac{4}{(e^x+e^{-x})^2}$$



      Second step : Factor $e^{-x}$ in the denominator :



      $$(e^x+e^{-x})^2=(e^{-x}(e^{2x}-1))^2=e^{-2x}(e^{2x}-1)^2$$



      We have obtained a $(e^{2x}-1)^2$ as a factor in the denominator.



      Getting rid of $e^{-2x}$ in the denominator by transforming it into $e^{2x}$ in the numerator, explains the final result with $4e^{2x}$ in the numerator.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Mar 16 at 10:50

























      answered Mar 16 at 6:21









      Jean MarieJean Marie

      31k42255




      31k42255








      • 1




        $begingroup$
        thank you, it was a very nice explanation. I understand it now.
        $endgroup$
        – Electric
        Mar 16 at 6:31














      • 1




        $begingroup$
        thank you, it was a very nice explanation. I understand it now.
        $endgroup$
        – Electric
        Mar 16 at 6:31








      1




      1




      $begingroup$
      thank you, it was a very nice explanation. I understand it now.
      $endgroup$
      – Electric
      Mar 16 at 6:31




      $begingroup$
      thank you, it was a very nice explanation. I understand it now.
      $endgroup$
      – Electric
      Mar 16 at 6:31











      0












      $begingroup$

      Hint: Use that $$a^2-b^2=(a-b)(a+b)$$






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        Hint: Use that $$a^2-b^2=(a-b)(a+b)$$






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          Hint: Use that $$a^2-b^2=(a-b)(a+b)$$






          share|cite|improve this answer









          $endgroup$



          Hint: Use that $$a^2-b^2=(a-b)(a+b)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 16 at 6:08









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          78.2k42867




          78.2k42867























              0












              $begingroup$

              So it's not even about derivative, as your question is that why:



              $$frac{a - b}{a} ne -b, a>0$$



              Well, they are just not equal, because in general $a-b+ab ne 0$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                So it's not even about derivative, as your question is that why:



                $$frac{a - b}{a} ne -b, a>0$$



                Well, they are just not equal, because in general $a-b+ab ne 0$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  So it's not even about derivative, as your question is that why:



                  $$frac{a - b}{a} ne -b, a>0$$



                  Well, they are just not equal, because in general $a-b+ab ne 0$






                  share|cite|improve this answer









                  $endgroup$



                  So it's not even about derivative, as your question is that why:



                  $$frac{a - b}{a} ne -b, a>0$$



                  Well, they are just not equal, because in general $a-b+ab ne 0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 16 at 6:15









                  Yujie ZhaYujie Zha

                  6,98611729




                  6,98611729






























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