Finding the 6th Derivative of a function of $sin(x)$












-1














I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:



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



The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:



$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$



I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question




















  • 1




    THIS might be of interest, especially with high-order derivatives.
    – Mark Viola
    Dec 9 '16 at 4:24
















-1














I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:



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



The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:



$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$



I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question




















  • 1




    THIS might be of interest, especially with high-order derivatives.
    – Mark Viola
    Dec 9 '16 at 4:24














-1












-1








-1







I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:



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



The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:



$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$



I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.






calculus sequences-and-series derivatives taylor-expansion







share|cite|improve this question















I am currently studying Maclaurin Series, and have come across a question that has me stumped. The function is:



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



The question asks me to find the 6th derivative of this function evaluated at $0$. Now I am trying to solve this with a Maclaurin Expansion of $sin(x)$, but I get:



$$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}}$$



I am struggling to find the 6th derivative, as the coefficient of $x^6$ is not obvious.






calculus sequences-and-series derivatives taylor-expansion




calculus sequences-and-series derivatives taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '16 at 5:22







user223391

















asked Dec 9 '16 at 4:21









Felicio Grande

499619




499619








  • 1




    THIS might be of interest, especially with high-order derivatives.
    – Mark Viola
    Dec 9 '16 at 4:24














  • 1




    THIS might be of interest, especially with high-order derivatives.
    – Mark Viola
    Dec 9 '16 at 4:24








1




1




THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24




THIS might be of interest, especially with high-order derivatives.
– Mark Viola
Dec 9 '16 at 4:24










3 Answers
3






active

oldest

votes


















3















You're on the right track.




Continuing, we see that



$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$




Therefore, the sixth derivative is $-(5/6)6!=-600$.







share|cite|improve this answer























  • Would the cowardly down voter care to comment??
    – Mark Viola
    Dec 9 '16 at 5:24










  • Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
    – Mark Viola
    Dec 9 '16 at 5:32










  • Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
    – Claude Leibovici
    Dec 9 '16 at 5:57






  • 1




    @ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
    – Mark Viola
    Dec 9 '16 at 6:18






  • 1




    @copper.hat Thank you! Much appreciative.
    – Mark Viola
    Dec 9 '16 at 6:51



















3














You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore




  • the sixth derivative is $-frac 5 6times 6!=-600$

  • the eighth derivative is $frac 2 3times 8!=26880$

  • the tenth derivative is $-frac {61}{120}times 10!=-1844640$






share|cite|improve this answer





















  • Tenacious! (+1)
    – Mark Viola
    Dec 9 '16 at 5:25










  • This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
    – Felicio Grande
    Dec 9 '16 at 5:26










  • @FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
    – Claude Leibovici
    Dec 9 '16 at 5:55





















2














Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$






share|cite|improve this answer





















  • @Dr.MV I think this is enough of a hint for now
    – user392576
    Dec 9 '16 at 4:28










  • @Dr.MV Of course it's correct. I already took that into account.
    – user392576
    Dec 9 '16 at 4:33











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









3















You're on the right track.




Continuing, we see that



$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$




Therefore, the sixth derivative is $-(5/6)6!=-600$.







share|cite|improve this answer























  • Would the cowardly down voter care to comment??
    – Mark Viola
    Dec 9 '16 at 5:24










  • Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
    – Mark Viola
    Dec 9 '16 at 5:32










  • Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
    – Claude Leibovici
    Dec 9 '16 at 5:57






  • 1




    @ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
    – Mark Viola
    Dec 9 '16 at 6:18






  • 1




    @copper.hat Thank you! Much appreciative.
    – Mark Viola
    Dec 9 '16 at 6:51
















3















You're on the right track.




Continuing, we see that



$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$




Therefore, the sixth derivative is $-(5/6)6!=-600$.







share|cite|improve this answer























  • Would the cowardly down voter care to comment??
    – Mark Viola
    Dec 9 '16 at 5:24










  • Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
    – Mark Viola
    Dec 9 '16 at 5:32










  • Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
    – Claude Leibovici
    Dec 9 '16 at 5:57






  • 1




    @ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
    – Mark Viola
    Dec 9 '16 at 6:18






  • 1




    @copper.hat Thank you! Much appreciative.
    – Mark Viola
    Dec 9 '16 at 6:51














3












3








3







You're on the right track.




Continuing, we see that



$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$




Therefore, the sixth derivative is $-(5/6)6!=-600$.







share|cite|improve this answer















You're on the right track.




Continuing, we see that



$$begin{align}
frac{1}{1+sin(x^2)}&=frac{1}{1+left(x^2-frac16x^6+Oleft(x^{10}right)right)}\\
&=1-left(x^2-frac16x^6+Oleft(x^{10}right)right)+left(x^2+Oleft(x^6right)right)^2-left(x^2+Oleft(x^6right)right)^3+O(x^8)\\
&=1-x^2+x^4-frac56x^6+O(x^8)
end{align}$$




Therefore, the sixth derivative is $-(5/6)6!=-600$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '16 at 4:43

























answered Dec 9 '16 at 4:37









Mark Viola

130k1273170




130k1273170












  • Would the cowardly down voter care to comment??
    – Mark Viola
    Dec 9 '16 at 5:24










  • Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
    – Mark Viola
    Dec 9 '16 at 5:32










  • Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
    – Claude Leibovici
    Dec 9 '16 at 5:57






  • 1




    @ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
    – Mark Viola
    Dec 9 '16 at 6:18






  • 1




    @copper.hat Thank you! Much appreciative.
    – Mark Viola
    Dec 9 '16 at 6:51


















  • Would the cowardly down voter care to comment??
    – Mark Viola
    Dec 9 '16 at 5:24










  • Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
    – Mark Viola
    Dec 9 '16 at 5:32










  • Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
    – Claude Leibovici
    Dec 9 '16 at 5:57






  • 1




    @ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
    – Mark Viola
    Dec 9 '16 at 6:18






  • 1




    @copper.hat Thank you! Much appreciative.
    – Mark Viola
    Dec 9 '16 at 6:51
















Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24




Would the cowardly down voter care to comment??
– Mark Viola
Dec 9 '16 at 5:24












Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32




Two cowardly down voters. Seriously? The answer is not useful? That is a dubious vote cast.
– Mark Viola
Dec 9 '16 at 5:32












Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57




Hey, Mark ! Good to see you. Please look at my comment to FelicioGrande; I never had the feeling that we compete (at the contrary !)
– Claude Leibovici
Dec 9 '16 at 5:57




1




1




@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18




@ClaudeLeibovici Hello my friend. I hope that you're enjoying this holiday season! And I believe that we support each other in bringing the best answers we can.
– Mark Viola
Dec 9 '16 at 6:18




1




1




@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51




@copper.hat Thank you! Much appreciative.
– Mark Viola
Dec 9 '16 at 6:51











3














You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore




  • the sixth derivative is $-frac 5 6times 6!=-600$

  • the eighth derivative is $frac 2 3times 8!=26880$

  • the tenth derivative is $-frac {61}{120}times 10!=-1844640$






share|cite|improve this answer





















  • Tenacious! (+1)
    – Mark Viola
    Dec 9 '16 at 5:25










  • This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
    – Felicio Grande
    Dec 9 '16 at 5:26










  • @FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
    – Claude Leibovici
    Dec 9 '16 at 5:55


















3














You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore




  • the sixth derivative is $-frac 5 6times 6!=-600$

  • the eighth derivative is $frac 2 3times 8!=26880$

  • the tenth derivative is $-frac {61}{120}times 10!=-1844640$






share|cite|improve this answer





















  • Tenacious! (+1)
    – Mark Viola
    Dec 9 '16 at 5:25










  • This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
    – Felicio Grande
    Dec 9 '16 at 5:26










  • @FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
    – Claude Leibovici
    Dec 9 '16 at 5:55
















3












3








3






You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore




  • the sixth derivative is $-frac 5 6times 6!=-600$

  • the eighth derivative is $frac 2 3times 8!=26880$

  • the tenth derivative is $-frac {61}{120}times 10!=-1844640$






share|cite|improve this answer












You properly wrote $$frac{1}{1+sin(x^2)} = frac{1}{1+x^2-frac{x^6}{6!}+frac{x^{10}}{5!}+Oleft(x^{12}right)}$$ Now perform the long division (by increasing powers of $x$) to get $$frac{1}{1+sin(x^2)} = 1-x^2+x^4-frac{5 x^6}{6}+frac{2 x^8}{3}-frac{61
x^{10}}{120}+Oleft(x^{12}right)$$ Therefore




  • the sixth derivative is $-frac 5 6times 6!=-600$

  • the eighth derivative is $frac 2 3times 8!=26880$

  • the tenth derivative is $-frac {61}{120}times 10!=-1844640$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '16 at 5:20









Claude Leibovici

118k1156132




118k1156132












  • Tenacious! (+1)
    – Mark Viola
    Dec 9 '16 at 5:25










  • This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
    – Felicio Grande
    Dec 9 '16 at 5:26










  • @FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
    – Claude Leibovici
    Dec 9 '16 at 5:55




















  • Tenacious! (+1)
    – Mark Viola
    Dec 9 '16 at 5:25










  • This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
    – Felicio Grande
    Dec 9 '16 at 5:26










  • @FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
    – Claude Leibovici
    Dec 9 '16 at 5:55


















Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25




Tenacious! (+1)
– Mark Viola
Dec 9 '16 at 5:25












This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26




This comment is great (+1), but Dr.MV had answered is first, thus my decision to give the checkmark to him. But thank you for this great answer!
– Felicio Grande
Dec 9 '16 at 5:26












@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55






@FelicioGrande. You are perfectly correct ! In any manner, I am not competing with my good friend Dr. MV. I was just proposing a way starting from what you wrote. Cheers.
– Claude Leibovici
Dec 9 '16 at 5:55













2














Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$






share|cite|improve this answer





















  • @Dr.MV I think this is enough of a hint for now
    – user392576
    Dec 9 '16 at 4:28










  • @Dr.MV Of course it's correct. I already took that into account.
    – user392576
    Dec 9 '16 at 4:33
















2














Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$






share|cite|improve this answer





















  • @Dr.MV I think this is enough of a hint for now
    – user392576
    Dec 9 '16 at 4:28










  • @Dr.MV Of course it's correct. I already took that into account.
    – user392576
    Dec 9 '16 at 4:33














2












2








2






Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$






share|cite|improve this answer












Use the geometric series i.e. $$frac{1}{1 + sin(x^2)} = 1 + Big( -x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big) + Big( - x^2 + frac{x^6}{3!} - frac{x^{10}}{5!} + ... Big)^2 + ...$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '16 at 4:25









user392576

1394




1394












  • @Dr.MV I think this is enough of a hint for now
    – user392576
    Dec 9 '16 at 4:28










  • @Dr.MV Of course it's correct. I already took that into account.
    – user392576
    Dec 9 '16 at 4:33


















  • @Dr.MV I think this is enough of a hint for now
    – user392576
    Dec 9 '16 at 4:28










  • @Dr.MV Of course it's correct. I already took that into account.
    – user392576
    Dec 9 '16 at 4:33
















@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28




@Dr.MV I think this is enough of a hint for now
– user392576
Dec 9 '16 at 4:28












@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33




@Dr.MV Of course it's correct. I already took that into account.
– user392576
Dec 9 '16 at 4:33


















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