Proof of quotient rule $(fracfg)'(x_0)=fracf'(x_0)g(x_0)-f(x_0)g'(x_0)g^2(x_0)$

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$$left(fracfgright)'(x_0)=fracf'(x_0)g(x_0)-f(x_0)g'(x_0)g^2(x_0)$$



So, $frac1g.f=fracfg$, then $$fracfg'(x_0)=fracf(x)frac1g(x)-f(x_0)frac1g(x_0)x-x_0=f(x)fracfrac1g(x)-frac1g(x_0)x-x_0-frac1g(x_0)fracf(x)-f(x_0)x-x_0$$ I took the limit of everything as $x to x_0$ $$=f(x_0)frac1g'(x_0)-frac1g(x_0)f'(x_0)=fracf(x_0)g'(x_0)-fracf'(x_0)g(x_0)=fracg(x_0)f(x_0)-f'(x_0)g'(x_0)g'(x_0)g(x_0)$$ which clearly it's fake.



Where did I go wrong?
Thank you!






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    up vote
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    $$left(fracfgright)'(x_0)=fracf'(x_0)g(x_0)-f(x_0)g'(x_0)g^2(x_0)$$



    So, $frac1g.f=fracfg$, then $$fracfg'(x_0)=fracf(x)frac1g(x)-f(x_0)frac1g(x_0)x-x_0=f(x)fracfrac1g(x)-frac1g(x_0)x-x_0-frac1g(x_0)fracf(x)-f(x_0)x-x_0$$ I took the limit of everything as $x to x_0$ $$=f(x_0)frac1g'(x_0)-frac1g(x_0)f'(x_0)=fracf(x_0)g'(x_0)-fracf'(x_0)g(x_0)=fracg(x_0)f(x_0)-f'(x_0)g'(x_0)g'(x_0)g(x_0)$$ which clearly it's fake.



    Where did I go wrong?
    Thank you!






    calculus real-analysis derivatives proof-writing







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite









      up vote
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      down vote

      favorite











      $$left(fracfgright)'(x_0)=fracf'(x_0)g(x_0)-f(x_0)g'(x_0)g^2(x_0)$$



      So, $frac1g.f=fracfg$, then $$fracfg'(x_0)=fracf(x)frac1g(x)-f(x_0)frac1g(x_0)x-x_0=f(x)fracfrac1g(x)-frac1g(x_0)x-x_0-frac1g(x_0)fracf(x)-f(x_0)x-x_0$$ I took the limit of everything as $x to x_0$ $$=f(x_0)frac1g'(x_0)-frac1g(x_0)f'(x_0)=fracf(x_0)g'(x_0)-fracf'(x_0)g(x_0)=fracg(x_0)f(x_0)-f'(x_0)g'(x_0)g'(x_0)g(x_0)$$ which clearly it's fake.



      Where did I go wrong?
      Thank you!






      calculus real-analysis derivatives proof-writing







      share|cite|improve this question















      $$left(fracfgright)'(x_0)=fracf'(x_0)g(x_0)-f(x_0)g'(x_0)g^2(x_0)$$



      So, $frac1g.f=fracfg$, then $$fracfg'(x_0)=fracf(x)frac1g(x)-f(x_0)frac1g(x_0)x-x_0=f(x)fracfrac1g(x)-frac1g(x_0)x-x_0-frac1g(x_0)fracf(x)-f(x_0)x-x_0$$ I took the limit of everything as $x to x_0$ $$=f(x_0)frac1g'(x_0)-frac1g(x_0)f'(x_0)=fracf(x_0)g'(x_0)-fracf'(x_0)g(x_0)=fracg(x_0)f(x_0)-f'(x_0)g'(x_0)g'(x_0)g(x_0)$$ which clearly it's fake.



      Where did I go wrong?
      Thank you!






      calculus real-analysis derivatives proof-writing




      calculus real-analysis derivatives proof-writing






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      edited 3 mins ago









      amWhy

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      asked 32 mins ago









      parishilton

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          You are close. Note that instead of cancelling the term $fracfracf(x)g(x_0)x-x_0$ you actually subtract it twice. Second, note that the $$lim_x rightarrow x_0 fracfrac1g(x) - frac1g(x_0)x-x_0 = left(frac1g(x_0)right)'neq frac1g'(x_0).$$



          Instead of doing it this way, I like to prove the quotient rule using the product rule: $fracfg = f cdot g^-1$. $fracddx (fg^-1) = f'g^-1 - fg'g^-2 = fracf'g - fg'g^2.$






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          • Thank you!! I got it right now. I actually think your method is easier haha
            – parishilton
            4 mins ago

















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          $(1/g)'neq 1/g'$ this is where you made the mistake.






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            2 Answers
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            up vote
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            accepted










            You are close. Note that instead of cancelling the term $fracfracf(x)g(x_0)x-x_0$ you actually subtract it twice. Second, note that the $$lim_x rightarrow x_0 fracfrac1g(x) - frac1g(x_0)x-x_0 = left(frac1g(x_0)right)'neq frac1g'(x_0).$$



            Instead of doing it this way, I like to prove the quotient rule using the product rule: $fracfg = f cdot g^-1$. $fracddx (fg^-1) = f'g^-1 - fg'g^-2 = fracf'g - fg'g^2.$






            share|cite|improve this answer




















            • Thank you!! I got it right now. I actually think your method is easier haha
              – parishilton
              4 mins ago














            up vote
            3
            down vote



            accepted










            You are close. Note that instead of cancelling the term $fracfracf(x)g(x_0)x-x_0$ you actually subtract it twice. Second, note that the $$lim_x rightarrow x_0 fracfrac1g(x) - frac1g(x_0)x-x_0 = left(frac1g(x_0)right)'neq frac1g'(x_0).$$



            Instead of doing it this way, I like to prove the quotient rule using the product rule: $fracfg = f cdot g^-1$. $fracddx (fg^-1) = f'g^-1 - fg'g^-2 = fracf'g - fg'g^2.$






            share|cite|improve this answer




















            • Thank you!! I got it right now. I actually think your method is easier haha
              – parishilton
              4 mins ago












            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            You are close. Note that instead of cancelling the term $fracfracf(x)g(x_0)x-x_0$ you actually subtract it twice. Second, note that the $$lim_x rightarrow x_0 fracfrac1g(x) - frac1g(x_0)x-x_0 = left(frac1g(x_0)right)'neq frac1g'(x_0).$$



            Instead of doing it this way, I like to prove the quotient rule using the product rule: $fracfg = f cdot g^-1$. $fracddx (fg^-1) = f'g^-1 - fg'g^-2 = fracf'g - fg'g^2.$






            share|cite|improve this answer












            You are close. Note that instead of cancelling the term $fracfracf(x)g(x_0)x-x_0$ you actually subtract it twice. Second, note that the $$lim_x rightarrow x_0 fracfrac1g(x) - frac1g(x_0)x-x_0 = left(frac1g(x_0)right)'neq frac1g'(x_0).$$



            Instead of doing it this way, I like to prove the quotient rule using the product rule: $fracfg = f cdot g^-1$. $fracddx (fg^-1) = f'g^-1 - fg'g^-2 = fracf'g - fg'g^2.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 15 mins ago









            Neeyanth Kopparapu

            464




            464











            • Thank you!! I got it right now. I actually think your method is easier haha
              – parishilton
              4 mins ago
















            • Thank you!! I got it right now. I actually think your method is easier haha
              – parishilton
              4 mins ago















            Thank you!! I got it right now. I actually think your method is easier haha
            – parishilton
            4 mins ago




            Thank you!! I got it right now. I actually think your method is easier haha
            – parishilton
            4 mins ago










            up vote
            3
            down vote













            $(1/g)'neq 1/g'$ this is where you made the mistake.






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              up vote
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              $(1/g)'neq 1/g'$ this is where you made the mistake.






              share|cite|improve this answer






















                up vote
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                up vote
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                $(1/g)'neq 1/g'$ this is where you made the mistake.






                share|cite|improve this answer












                $(1/g)'neq 1/g'$ this is where you made the mistake.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 20 mins ago









                Singh

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