Find the derivative of the function $(y^2-1)^2/(y^2+1)$ [closed]

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How do I solve this? I tried using the quotient and the chain rule but I can't seem to get the correct answer.






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closed as off-topic by José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh Sep 2 at 12:37


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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    Welcome to MSE! Could you show us your steps so that we can identify your error?
    – Sobi
    Sep 2 at 11:15










  • derivative with respect to what? $y$?$x$?
    – Pablo
    Sep 2 at 12:02














up vote
0
down vote

favorite












How do I solve this? I tried using the quotient and the chain rule but I can't seem to get the correct answer.






derivatives







share|cite|improve this question















closed as off-topic by José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh Sep 2 at 12:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    Welcome to MSE! Could you show us your steps so that we can identify your error?
    – Sobi
    Sep 2 at 11:15










  • derivative with respect to what? $y$?$x$?
    – Pablo
    Sep 2 at 12:02












up vote
0
down vote

favorite









up vote
0
down vote

favorite











How do I solve this? I tried using the quotient and the chain rule but I can't seem to get the correct answer.






derivatives







share|cite|improve this question















How do I solve this? I tried using the quotient and the chain rule but I can't seem to get the correct answer.






derivatives




derivatives






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edited Sep 2 at 11:22









Henning Makholm

231k16297529




231k16297529










asked Sep 2 at 11:14









Joseph Anthony

103




103




closed as off-topic by José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh Sep 2 at 12:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh Sep 2 at 12:37


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Henning Makholm, Lord Shark the Unknown, Jose Arnaldo Bebita Dris, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 5




    Welcome to MSE! Could you show us your steps so that we can identify your error?
    – Sobi
    Sep 2 at 11:15










  • derivative with respect to what? $y$?$x$?
    – Pablo
    Sep 2 at 12:02












  • 5




    Welcome to MSE! Could you show us your steps so that we can identify your error?
    – Sobi
    Sep 2 at 11:15










  • derivative with respect to what? $y$?$x$?
    – Pablo
    Sep 2 at 12:02







5




5




Welcome to MSE! Could you show us your steps so that we can identify your error?
– Sobi
Sep 2 at 11:15




Welcome to MSE! Could you show us your steps so that we can identify your error?
– Sobi
Sep 2 at 11:15












derivative with respect to what? $y$?$x$?
– Pablo
Sep 2 at 12:02




derivative with respect to what? $y$?$x$?
– Pablo
Sep 2 at 12:02










3 Answers
3






active

oldest

votes

















up vote
0
down vote



accepted










You can set $;f(y)=dfrac(y^2-1)^2y^2+1$ and use logarithmic differentiation:
beginalign
fracf'(y)f(y)&=2,frac(y^2-1)'y^2-1-frac(y^2+1)'y2+1=2,frac2yy^2-1-frac2yy^2+1=2ybiggl(frac2y^2-1-frac1y^2+1biggr)=frac2y(y^2+5)(y^2-1)(y^2+1)
endalign
whence
$$ f'(y)=fracf'(y)f(y),f(y)=frac2y(y^2+5)(y^2-1)(y^2+1)^2. $$






share|cite|improve this answer



























    up vote
    1
    down vote













    HINT:



    You can simplify the chain rule using $$(y^2-1)^2=(y^2+1)-2^2=?$$






    share|cite|improve this answer






















    • Won't some of the remaining terms still need chain rule? (although simpler)
      – GoodDeeds
      Sep 2 at 11:21










    • @GoodDeeds, Agreed
      – lab bhattacharjee
      Sep 2 at 11:34

















    up vote
    0
    down vote













    Let:
    $$p=(y^2-1)^2to p'=2(2y)(y^2-1)=4y(y^2-1) text [chain rule]$$
    $$q=y^2+1 to q'=2y text [implicit]$$
    Then the quotient rule:
    $$bigg(frac pqbigg)'=fracp'q-q'pq^2$$
    See if you can take this forward






    share|cite|improve this answer




















    • I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
      – Joseph Anthony
      Sep 2 at 11:39










    • Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
      – Rhys Hughes
      Sep 2 at 11:54


















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote



    accepted










    You can set $;f(y)=dfrac(y^2-1)^2y^2+1$ and use logarithmic differentiation:
    beginalign
    fracf'(y)f(y)&=2,frac(y^2-1)'y^2-1-frac(y^2+1)'y2+1=2,frac2yy^2-1-frac2yy^2+1=2ybiggl(frac2y^2-1-frac1y^2+1biggr)=frac2y(y^2+5)(y^2-1)(y^2+1)
    endalign
    whence
    $$ f'(y)=fracf'(y)f(y),f(y)=frac2y(y^2+5)(y^2-1)(y^2+1)^2. $$






    share|cite|improve this answer
























      up vote
      0
      down vote



      accepted










      You can set $;f(y)=dfrac(y^2-1)^2y^2+1$ and use logarithmic differentiation:
      beginalign
      fracf'(y)f(y)&=2,frac(y^2-1)'y^2-1-frac(y^2+1)'y2+1=2,frac2yy^2-1-frac2yy^2+1=2ybiggl(frac2y^2-1-frac1y^2+1biggr)=frac2y(y^2+5)(y^2-1)(y^2+1)
      endalign
      whence
      $$ f'(y)=fracf'(y)f(y),f(y)=frac2y(y^2+5)(y^2-1)(y^2+1)^2. $$






      share|cite|improve this answer






















        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        You can set $;f(y)=dfrac(y^2-1)^2y^2+1$ and use logarithmic differentiation:
        beginalign
        fracf'(y)f(y)&=2,frac(y^2-1)'y^2-1-frac(y^2+1)'y2+1=2,frac2yy^2-1-frac2yy^2+1=2ybiggl(frac2y^2-1-frac1y^2+1biggr)=frac2y(y^2+5)(y^2-1)(y^2+1)
        endalign
        whence
        $$ f'(y)=fracf'(y)f(y),f(y)=frac2y(y^2+5)(y^2-1)(y^2+1)^2. $$






        share|cite|improve this answer












        You can set $;f(y)=dfrac(y^2-1)^2y^2+1$ and use logarithmic differentiation:
        beginalign
        fracf'(y)f(y)&=2,frac(y^2-1)'y^2-1-frac(y^2+1)'y2+1=2,frac2yy^2-1-frac2yy^2+1=2ybiggl(frac2y^2-1-frac1y^2+1biggr)=frac2y(y^2+5)(y^2-1)(y^2+1)
        endalign
        whence
        $$ f'(y)=fracf'(y)f(y),f(y)=frac2y(y^2+5)(y^2-1)(y^2+1)^2. $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 2 at 12:03









        Bernard

        112k635104




        112k635104




















            up vote
            1
            down vote













            HINT:



            You can simplify the chain rule using $$(y^2-1)^2=(y^2+1)-2^2=?$$






            share|cite|improve this answer






















            • Won't some of the remaining terms still need chain rule? (although simpler)
              – GoodDeeds
              Sep 2 at 11:21










            • @GoodDeeds, Agreed
              – lab bhattacharjee
              Sep 2 at 11:34














            up vote
            1
            down vote













            HINT:



            You can simplify the chain rule using $$(y^2-1)^2=(y^2+1)-2^2=?$$






            share|cite|improve this answer






















            • Won't some of the remaining terms still need chain rule? (although simpler)
              – GoodDeeds
              Sep 2 at 11:21










            • @GoodDeeds, Agreed
              – lab bhattacharjee
              Sep 2 at 11:34












            up vote
            1
            down vote










            up vote
            1
            down vote









            HINT:



            You can simplify the chain rule using $$(y^2-1)^2=(y^2+1)-2^2=?$$






            share|cite|improve this answer














            HINT:



            You can simplify the chain rule using $$(y^2-1)^2=(y^2+1)-2^2=?$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Sep 2 at 11:34

























            answered Sep 2 at 11:19









            lab bhattacharjee

            216k14153265




            216k14153265











            • Won't some of the remaining terms still need chain rule? (although simpler)
              – GoodDeeds
              Sep 2 at 11:21










            • @GoodDeeds, Agreed
              – lab bhattacharjee
              Sep 2 at 11:34
















            • Won't some of the remaining terms still need chain rule? (although simpler)
              – GoodDeeds
              Sep 2 at 11:21










            • @GoodDeeds, Agreed
              – lab bhattacharjee
              Sep 2 at 11:34















            Won't some of the remaining terms still need chain rule? (although simpler)
            – GoodDeeds
            Sep 2 at 11:21




            Won't some of the remaining terms still need chain rule? (although simpler)
            – GoodDeeds
            Sep 2 at 11:21












            @GoodDeeds, Agreed
            – lab bhattacharjee
            Sep 2 at 11:34




            @GoodDeeds, Agreed
            – lab bhattacharjee
            Sep 2 at 11:34










            up vote
            0
            down vote













            Let:
            $$p=(y^2-1)^2to p'=2(2y)(y^2-1)=4y(y^2-1) text [chain rule]$$
            $$q=y^2+1 to q'=2y text [implicit]$$
            Then the quotient rule:
            $$bigg(frac pqbigg)'=fracp'q-q'pq^2$$
            See if you can take this forward






            share|cite|improve this answer




















            • I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
              – Joseph Anthony
              Sep 2 at 11:39










            • Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
              – Rhys Hughes
              Sep 2 at 11:54















            up vote
            0
            down vote













            Let:
            $$p=(y^2-1)^2to p'=2(2y)(y^2-1)=4y(y^2-1) text [chain rule]$$
            $$q=y^2+1 to q'=2y text [implicit]$$
            Then the quotient rule:
            $$bigg(frac pqbigg)'=fracp'q-q'pq^2$$
            See if you can take this forward






            share|cite|improve this answer




















            • I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
              – Joseph Anthony
              Sep 2 at 11:39










            • Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
              – Rhys Hughes
              Sep 2 at 11:54













            up vote
            0
            down vote










            up vote
            0
            down vote









            Let:
            $$p=(y^2-1)^2to p'=2(2y)(y^2-1)=4y(y^2-1) text [chain rule]$$
            $$q=y^2+1 to q'=2y text [implicit]$$
            Then the quotient rule:
            $$bigg(frac pqbigg)'=fracp'q-q'pq^2$$
            See if you can take this forward






            share|cite|improve this answer












            Let:
            $$p=(y^2-1)^2to p'=2(2y)(y^2-1)=4y(y^2-1) text [chain rule]$$
            $$q=y^2+1 to q'=2y text [implicit]$$
            Then the quotient rule:
            $$bigg(frac pqbigg)'=fracp'q-q'pq^2$$
            See if you can take this forward







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Sep 2 at 11:19









            Rhys Hughes

            4,1231227




            4,1231227











            • I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
              – Joseph Anthony
              Sep 2 at 11:39










            • Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
              – Rhys Hughes
              Sep 2 at 11:54

















            • I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
              – Joseph Anthony
              Sep 2 at 11:39










            • Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
              – Rhys Hughes
              Sep 2 at 11:54
















            I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
            – Joseph Anthony
            Sep 2 at 11:39




            I can't simplify it, I get confused on what to do next when I substitute the value of p, q, p' and q' in the quotient rule
            – Joseph Anthony
            Sep 2 at 11:39












            Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
            – Rhys Hughes
            Sep 2 at 11:54





            Plugging in the values we get: $$bigg(frac(y^2-1)^2y^2+1bigg)'=frac4y(y^2-1)(y^2+1)-2y(y^2-1)^2(y^2+1)^2$$ You can then take out common factors for $$frac2y(y^2-1)[2(y^2+1)-(y^2-1)](y^2+1)^2$$ Go from there.
            – Rhys Hughes
            Sep 2 at 11:54



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