Applying product rule of differentiation












1












$begingroup$


I have been given this equation in a past exam paper, in which I am solving for eigenfunctions & eigenvalues.



The product rule has been applied to this section in the solution but I cannot work out which should be 'f' and which should be 'g'.



The result is : after dividing through by , which I understand but how do I perform the product rule on my first equation?



Thanks






calculus ordinary-differential-equations







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    1












    $begingroup$


    I have been given this equation in a past exam paper, in which I am solving for eigenfunctions & eigenvalues.



    The product rule has been applied to this section in the solution but I cannot work out which should be 'f' and which should be 'g'.



    The result is : after dividing through by , which I understand but how do I perform the product rule on my first equation?



    Thanks






    calculus ordinary-differential-equations







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I have been given this equation in a past exam paper, in which I am solving for eigenfunctions & eigenvalues.



      The product rule has been applied to this section in the solution but I cannot work out which should be 'f' and which should be 'g'.



      The result is : after dividing through by , which I understand but how do I perform the product rule on my first equation?



      Thanks






      calculus ordinary-differential-equations







      share|cite|improve this question









      $endgroup$




      I have been given this equation in a past exam paper, in which I am solving for eigenfunctions & eigenvalues.



      The product rule has been applied to this section in the solution but I cannot work out which should be 'f' and which should be 'g'.



      The result is : after dividing through by , which I understand but how do I perform the product rule on my first equation?



      Thanks






      calculus ordinary-differential-equations




      calculus ordinary-differential-equations






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      share|cite|improve this question











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      asked Jan 6 at 13:50









      JonnyWhite13JonnyWhite13

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          2 Answers
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          I would write $$-frac{d}{dx}(e^{-2x}frac{dy}{dx})=-frac{d}{dx}e^{-2x}frac{dy}{dx}-e^{-2x}frac{d}{dx}(frac{dy}{dx})$$ and this is $$2e^{-2x}frac{dy}{dx}-e^{-2x}frac{d^2y}{dx^2}$$






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          • $begingroup$
            Yes that makes perfect sense, thanks!
            $endgroup$
            – JonnyWhite13
            Jan 6 at 14:00



















          3












          $begingroup$

          Just use $(fg)’ = f’g+fg’$ where $f = e^{-2x}$ and $g = frac{dy}{dx}$.



          $$f’ = frac{d}{dx} e^{-2x} = -2e^{-2x}$$



          $$g’ = frac{dfrac{dy}{dx}}{dx} = frac{d^2y}{dx^2}$$



          So, considering there is a negative sign as well, putting everything together, you get



          $$-left[-2e^{-2x}cdotfrac{dy}{dx}+e^{-2x}cdotfrac{d^2y}{dx^2}right] = 2e^{-2x}cdotfrac{dy}{dx}-e^{-2x}cdotfrac{d^2y}{dx^2}$$






          share|cite|improve this answer











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          • $begingroup$
            That also makes lots more sense.
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58










          • $begingroup$
            Thank you very much
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58











          Your Answer





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          2 Answers
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          2 Answers
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          active

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          3












          $begingroup$

          I would write $$-frac{d}{dx}(e^{-2x}frac{dy}{dx})=-frac{d}{dx}e^{-2x}frac{dy}{dx}-e^{-2x}frac{d}{dx}(frac{dy}{dx})$$ and this is $$2e^{-2x}frac{dy}{dx}-e^{-2x}frac{d^2y}{dx^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes that makes perfect sense, thanks!
            $endgroup$
            – JonnyWhite13
            Jan 6 at 14:00
















          3












          $begingroup$

          I would write $$-frac{d}{dx}(e^{-2x}frac{dy}{dx})=-frac{d}{dx}e^{-2x}frac{dy}{dx}-e^{-2x}frac{d}{dx}(frac{dy}{dx})$$ and this is $$2e^{-2x}frac{dy}{dx}-e^{-2x}frac{d^2y}{dx^2}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes that makes perfect sense, thanks!
            $endgroup$
            – JonnyWhite13
            Jan 6 at 14:00














          3












          3








          3





          $begingroup$

          I would write $$-frac{d}{dx}(e^{-2x}frac{dy}{dx})=-frac{d}{dx}e^{-2x}frac{dy}{dx}-e^{-2x}frac{d}{dx}(frac{dy}{dx})$$ and this is $$2e^{-2x}frac{dy}{dx}-e^{-2x}frac{d^2y}{dx^2}$$






          share|cite|improve this answer









          $endgroup$



          I would write $$-frac{d}{dx}(e^{-2x}frac{dy}{dx})=-frac{d}{dx}e^{-2x}frac{dy}{dx}-e^{-2x}frac{d}{dx}(frac{dy}{dx})$$ and this is $$2e^{-2x}frac{dy}{dx}-e^{-2x}frac{d^2y}{dx^2}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 13:57









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          75.2k42865




          75.2k42865












          • $begingroup$
            Yes that makes perfect sense, thanks!
            $endgroup$
            – JonnyWhite13
            Jan 6 at 14:00


















          • $begingroup$
            Yes that makes perfect sense, thanks!
            $endgroup$
            – JonnyWhite13
            Jan 6 at 14:00
















          $begingroup$
          Yes that makes perfect sense, thanks!
          $endgroup$
          – JonnyWhite13
          Jan 6 at 14:00




          $begingroup$
          Yes that makes perfect sense, thanks!
          $endgroup$
          – JonnyWhite13
          Jan 6 at 14:00











          3












          $begingroup$

          Just use $(fg)’ = f’g+fg’$ where $f = e^{-2x}$ and $g = frac{dy}{dx}$.



          $$f’ = frac{d}{dx} e^{-2x} = -2e^{-2x}$$



          $$g’ = frac{dfrac{dy}{dx}}{dx} = frac{d^2y}{dx^2}$$



          So, considering there is a negative sign as well, putting everything together, you get



          $$-left[-2e^{-2x}cdotfrac{dy}{dx}+e^{-2x}cdotfrac{d^2y}{dx^2}right] = 2e^{-2x}cdotfrac{dy}{dx}-e^{-2x}cdotfrac{d^2y}{dx^2}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That also makes lots more sense.
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58










          • $begingroup$
            Thank you very much
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58
















          3












          $begingroup$

          Just use $(fg)’ = f’g+fg’$ where $f = e^{-2x}$ and $g = frac{dy}{dx}$.



          $$f’ = frac{d}{dx} e^{-2x} = -2e^{-2x}$$



          $$g’ = frac{dfrac{dy}{dx}}{dx} = frac{d^2y}{dx^2}$$



          So, considering there is a negative sign as well, putting everything together, you get



          $$-left[-2e^{-2x}cdotfrac{dy}{dx}+e^{-2x}cdotfrac{d^2y}{dx^2}right] = 2e^{-2x}cdotfrac{dy}{dx}-e^{-2x}cdotfrac{d^2y}{dx^2}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            That also makes lots more sense.
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58










          • $begingroup$
            Thank you very much
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58














          3












          3








          3





          $begingroup$

          Just use $(fg)’ = f’g+fg’$ where $f = e^{-2x}$ and $g = frac{dy}{dx}$.



          $$f’ = frac{d}{dx} e^{-2x} = -2e^{-2x}$$



          $$g’ = frac{dfrac{dy}{dx}}{dx} = frac{d^2y}{dx^2}$$



          So, considering there is a negative sign as well, putting everything together, you get



          $$-left[-2e^{-2x}cdotfrac{dy}{dx}+e^{-2x}cdotfrac{d^2y}{dx^2}right] = 2e^{-2x}cdotfrac{dy}{dx}-e^{-2x}cdotfrac{d^2y}{dx^2}$$






          share|cite|improve this answer











          $endgroup$



          Just use $(fg)’ = f’g+fg’$ where $f = e^{-2x}$ and $g = frac{dy}{dx}$.



          $$f’ = frac{d}{dx} e^{-2x} = -2e^{-2x}$$



          $$g’ = frac{dfrac{dy}{dx}}{dx} = frac{d^2y}{dx^2}$$



          So, considering there is a negative sign as well, putting everything together, you get



          $$-left[-2e^{-2x}cdotfrac{dy}{dx}+e^{-2x}cdotfrac{d^2y}{dx^2}right] = 2e^{-2x}cdotfrac{dy}{dx}-e^{-2x}cdotfrac{d^2y}{dx^2}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 21:32

























          answered Jan 6 at 14:04









          KM101KM101

          5,9951524




          5,9951524












          • $begingroup$
            That also makes lots more sense.
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58










          • $begingroup$
            Thank you very much
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58


















          • $begingroup$
            That also makes lots more sense.
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58










          • $begingroup$
            Thank you very much
            $endgroup$
            – JonnyWhite13
            Jan 6 at 17:58
















          $begingroup$
          That also makes lots more sense.
          $endgroup$
          – JonnyWhite13
          Jan 6 at 17:58




          $begingroup$
          That also makes lots more sense.
          $endgroup$
          – JonnyWhite13
          Jan 6 at 17:58












          $begingroup$
          Thank you very much
          $endgroup$
          – JonnyWhite13
          Jan 6 at 17:58




          $begingroup$
          Thank you very much
          $endgroup$
          – JonnyWhite13
          Jan 6 at 17:58


















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