What am I doing wrong finding the $n$-th derivative of $arcsin(x)$ and its value at $0$?












0














$f(x)=arcsin(x)$



$f'(x)=frac{1}{f^{-1}(x)}$



$f'(arcsin(x))=frac{1}{sin'x}=cos^{-1}x$



$(f^{(n-1)})'(x)=(cos^{-1}x)^{(n-1)}$



It's a complex function: $x$ itself, $cos x$, and $^{-1}$. Let's use the chain rule.



$(cos^{-1}x)^{(n-1)}=cos (x+frac{(n-1)pi}{2})cdot (-1 cdot -2 cdot cdots cdot (-1-n+1+1))cos^{-1-n+1}x=$



$=cos (x+frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!cdot cos^{-n} x$



$f^{(n)}(0)=cos (frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!$



If $n$ is even, then the cosine turns into zero, and the whole derivative becomes zero.



If $n$ is odd, then



$f^{(2k+1)}(0)=cos (frac{(2k)pi}{2})cdot (2k)!=(-1)^{k}cdot (2k)!$.



The book, however, gives $(1cdot 3cdot cdots cdot (2k-1))^2$ for the odd case.



Do I misunderstand something fundamentally?






calculus derivatives







share|cite|improve this question
























  • Your derivative is incorrect
    – Randall
    Dec 3 '18 at 15:36










  • The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
    – Ethan Bolker
    Dec 3 '18 at 15:38












  • $arcsin x ne frac{1}{sin x}$
    – Vasya
    Dec 3 '18 at 15:50










  • Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
    – B. Goddard
    Dec 3 '18 at 16:11
















0














$f(x)=arcsin(x)$



$f'(x)=frac{1}{f^{-1}(x)}$



$f'(arcsin(x))=frac{1}{sin'x}=cos^{-1}x$



$(f^{(n-1)})'(x)=(cos^{-1}x)^{(n-1)}$



It's a complex function: $x$ itself, $cos x$, and $^{-1}$. Let's use the chain rule.



$(cos^{-1}x)^{(n-1)}=cos (x+frac{(n-1)pi}{2})cdot (-1 cdot -2 cdot cdots cdot (-1-n+1+1))cos^{-1-n+1}x=$



$=cos (x+frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!cdot cos^{-n} x$



$f^{(n)}(0)=cos (frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!$



If $n$ is even, then the cosine turns into zero, and the whole derivative becomes zero.



If $n$ is odd, then



$f^{(2k+1)}(0)=cos (frac{(2k)pi}{2})cdot (2k)!=(-1)^{k}cdot (2k)!$.



The book, however, gives $(1cdot 3cdot cdots cdot (2k-1))^2$ for the odd case.



Do I misunderstand something fundamentally?






calculus derivatives







share|cite|improve this question
























  • Your derivative is incorrect
    – Randall
    Dec 3 '18 at 15:36










  • The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
    – Ethan Bolker
    Dec 3 '18 at 15:38












  • $arcsin x ne frac{1}{sin x}$
    – Vasya
    Dec 3 '18 at 15:50










  • Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
    – B. Goddard
    Dec 3 '18 at 16:11














0












0








0







$f(x)=arcsin(x)$



$f'(x)=frac{1}{f^{-1}(x)}$



$f'(arcsin(x))=frac{1}{sin'x}=cos^{-1}x$



$(f^{(n-1)})'(x)=(cos^{-1}x)^{(n-1)}$



It's a complex function: $x$ itself, $cos x$, and $^{-1}$. Let's use the chain rule.



$(cos^{-1}x)^{(n-1)}=cos (x+frac{(n-1)pi}{2})cdot (-1 cdot -2 cdot cdots cdot (-1-n+1+1))cos^{-1-n+1}x=$



$=cos (x+frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!cdot cos^{-n} x$



$f^{(n)}(0)=cos (frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!$



If $n$ is even, then the cosine turns into zero, and the whole derivative becomes zero.



If $n$ is odd, then



$f^{(2k+1)}(0)=cos (frac{(2k)pi}{2})cdot (2k)!=(-1)^{k}cdot (2k)!$.



The book, however, gives $(1cdot 3cdot cdots cdot (2k-1))^2$ for the odd case.



Do I misunderstand something fundamentally?






calculus derivatives







share|cite|improve this question















$f(x)=arcsin(x)$



$f'(x)=frac{1}{f^{-1}(x)}$



$f'(arcsin(x))=frac{1}{sin'x}=cos^{-1}x$



$(f^{(n-1)})'(x)=(cos^{-1}x)^{(n-1)}$



It's a complex function: $x$ itself, $cos x$, and $^{-1}$. Let's use the chain rule.



$(cos^{-1}x)^{(n-1)}=cos (x+frac{(n-1)pi}{2})cdot (-1 cdot -2 cdot cdots cdot (-1-n+1+1))cos^{-1-n+1}x=$



$=cos (x+frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!cdot cos^{-n} x$



$f^{(n)}(0)=cos (frac{(n-1)pi}{2})cdot (-1)^{n-1}cdot (n-1)!$



If $n$ is even, then the cosine turns into zero, and the whole derivative becomes zero.



If $n$ is odd, then



$f^{(2k+1)}(0)=cos (frac{(2k)pi}{2})cdot (2k)!=(-1)^{k}cdot (2k)!$.



The book, however, gives $(1cdot 3cdot cdots cdot (2k-1))^2$ for the odd case.



Do I misunderstand something fundamentally?






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 Dec 3 '18 at 15:39









Jneven

742320




742320










asked Dec 3 '18 at 15:34









fragileradius

297114




297114












  • Your derivative is incorrect
    – Randall
    Dec 3 '18 at 15:36










  • The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
    – Ethan Bolker
    Dec 3 '18 at 15:38












  • $arcsin x ne frac{1}{sin x}$
    – Vasya
    Dec 3 '18 at 15:50










  • Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
    – B. Goddard
    Dec 3 '18 at 16:11


















  • Your derivative is incorrect
    – Randall
    Dec 3 '18 at 15:36










  • The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
    – Ethan Bolker
    Dec 3 '18 at 15:38












  • $arcsin x ne frac{1}{sin x}$
    – Vasya
    Dec 3 '18 at 15:50










  • Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
    – B. Goddard
    Dec 3 '18 at 16:11
















Your derivative is incorrect
– Randall
Dec 3 '18 at 15:36




Your derivative is incorrect
– Randall
Dec 3 '18 at 15:36












The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
– Ethan Bolker
Dec 3 '18 at 15:38






The derivative of $arcsin x$ is $1/sqrt{1-x^2}$.
– Ethan Bolker
Dec 3 '18 at 15:38














$arcsin x ne frac{1}{sin x}$
– Vasya
Dec 3 '18 at 15:50




$arcsin x ne frac{1}{sin x}$
– Vasya
Dec 3 '18 at 15:50












Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
– B. Goddard
Dec 3 '18 at 16:11




Likely, the point of the exercise is that, rather than take any derivatives, you find the Maclauren series for $arcsin x$ (by, say, binomial theorem) and then just observe the values of the derivatives at $0$.
– B. Goddard
Dec 3 '18 at 16:11










2 Answers
2






active

oldest

votes


















2














You’re using the incorrect formula for the derivative of an inverse function.



You wanted to use



$$(f^{-1}(x))’ = frac{1}{f’(f^{-1}(x))}$$



Without using the formula, all you really need is the Chain Rule.



$$y = arcsin x iff sin y = x$$



$$frac{d}{dx} sin y = frac{d}{dx}x$$



$$frac{du}{dx} = frac{du}{dy}cdotfrac{dy}{dx} implies cos ycdotfrac{dy}{dx} = 1$$



$$frac{dy}{dx} = frac{1}{cos y} implies frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2y}}$$



$$frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2(arcsin x)}} = frac{1}{color{blue}{+}sqrt{1-x^2}}$$



Note the positive sign as $cos y$ is positive in the range of $arcsin x$, which is $big[-frac{pi}{2}, frac{pi}{2}big]$.






share|cite|improve this answer































    1














    Your way to proceed is really not clear, the standard way is as follows



    $$y=arcsin x implies sin y = x implies cos ycdot y'=1 implies y'=frac1{cos y}$$



    and since $cos y$ is positive in the range of $arcsin x$



    $$cos y= cos (arcsin x)=sqrt{1-sin^2(arcsin x)}=sqrt{1-x^2}$$






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024203%2fwhat-am-i-doing-wrong-finding-the-n-th-derivative-of-arcsinx-and-its-valu%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      You’re using the incorrect formula for the derivative of an inverse function.



      You wanted to use



      $$(f^{-1}(x))’ = frac{1}{f’(f^{-1}(x))}$$



      Without using the formula, all you really need is the Chain Rule.



      $$y = arcsin x iff sin y = x$$



      $$frac{d}{dx} sin y = frac{d}{dx}x$$



      $$frac{du}{dx} = frac{du}{dy}cdotfrac{dy}{dx} implies cos ycdotfrac{dy}{dx} = 1$$



      $$frac{dy}{dx} = frac{1}{cos y} implies frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2y}}$$



      $$frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2(arcsin x)}} = frac{1}{color{blue}{+}sqrt{1-x^2}}$$



      Note the positive sign as $cos y$ is positive in the range of $arcsin x$, which is $big[-frac{pi}{2}, frac{pi}{2}big]$.






      share|cite|improve this answer




























        2














        You’re using the incorrect formula for the derivative of an inverse function.



        You wanted to use



        $$(f^{-1}(x))’ = frac{1}{f’(f^{-1}(x))}$$



        Without using the formula, all you really need is the Chain Rule.



        $$y = arcsin x iff sin y = x$$



        $$frac{d}{dx} sin y = frac{d}{dx}x$$



        $$frac{du}{dx} = frac{du}{dy}cdotfrac{dy}{dx} implies cos ycdotfrac{dy}{dx} = 1$$



        $$frac{dy}{dx} = frac{1}{cos y} implies frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2y}}$$



        $$frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2(arcsin x)}} = frac{1}{color{blue}{+}sqrt{1-x^2}}$$



        Note the positive sign as $cos y$ is positive in the range of $arcsin x$, which is $big[-frac{pi}{2}, frac{pi}{2}big]$.






        share|cite|improve this answer


























          2












          2








          2






          You’re using the incorrect formula for the derivative of an inverse function.



          You wanted to use



          $$(f^{-1}(x))’ = frac{1}{f’(f^{-1}(x))}$$



          Without using the formula, all you really need is the Chain Rule.



          $$y = arcsin x iff sin y = x$$



          $$frac{d}{dx} sin y = frac{d}{dx}x$$



          $$frac{du}{dx} = frac{du}{dy}cdotfrac{dy}{dx} implies cos ycdotfrac{dy}{dx} = 1$$



          $$frac{dy}{dx} = frac{1}{cos y} implies frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2y}}$$



          $$frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2(arcsin x)}} = frac{1}{color{blue}{+}sqrt{1-x^2}}$$



          Note the positive sign as $cos y$ is positive in the range of $arcsin x$, which is $big[-frac{pi}{2}, frac{pi}{2}big]$.






          share|cite|improve this answer














          You’re using the incorrect formula for the derivative of an inverse function.



          You wanted to use



          $$(f^{-1}(x))’ = frac{1}{f’(f^{-1}(x))}$$



          Without using the formula, all you really need is the Chain Rule.



          $$y = arcsin x iff sin y = x$$



          $$frac{d}{dx} sin y = frac{d}{dx}x$$



          $$frac{du}{dx} = frac{du}{dy}cdotfrac{dy}{dx} implies cos ycdotfrac{dy}{dx} = 1$$



          $$frac{dy}{dx} = frac{1}{cos y} implies frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2y}}$$



          $$frac{dy}{dx} = frac{1}{color{blue}{+}sqrt{1-sin^2(arcsin x)}} = frac{1}{color{blue}{+}sqrt{1-x^2}}$$



          Note the positive sign as $cos y$ is positive in the range of $arcsin x$, which is $big[-frac{pi}{2}, frac{pi}{2}big]$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 3 '18 at 16:07

























          answered Dec 3 '18 at 16:00









          KM101

          5,0141423




          5,0141423























              1














              Your way to proceed is really not clear, the standard way is as follows



              $$y=arcsin x implies sin y = x implies cos ycdot y'=1 implies y'=frac1{cos y}$$



              and since $cos y$ is positive in the range of $arcsin x$



              $$cos y= cos (arcsin x)=sqrt{1-sin^2(arcsin x)}=sqrt{1-x^2}$$






              share|cite|improve this answer


























                1














                Your way to proceed is really not clear, the standard way is as follows



                $$y=arcsin x implies sin y = x implies cos ycdot y'=1 implies y'=frac1{cos y}$$



                and since $cos y$ is positive in the range of $arcsin x$



                $$cos y= cos (arcsin x)=sqrt{1-sin^2(arcsin x)}=sqrt{1-x^2}$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  Your way to proceed is really not clear, the standard way is as follows



                  $$y=arcsin x implies sin y = x implies cos ycdot y'=1 implies y'=frac1{cos y}$$



                  and since $cos y$ is positive in the range of $arcsin x$



                  $$cos y= cos (arcsin x)=sqrt{1-sin^2(arcsin x)}=sqrt{1-x^2}$$






                  share|cite|improve this answer












                  Your way to proceed is really not clear, the standard way is as follows



                  $$y=arcsin x implies sin y = x implies cos ycdot y'=1 implies y'=frac1{cos y}$$



                  and since $cos y$ is positive in the range of $arcsin x$



                  $$cos y= cos (arcsin x)=sqrt{1-sin^2(arcsin x)}=sqrt{1-x^2}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 3 '18 at 15:48









                  gimusi

                  1




                  1






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024203%2fwhat-am-i-doing-wrong-finding-the-n-th-derivative-of-arcsinx-and-its-valu%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      Berounka

                      Image
                      Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
                      Read more

                      Sphinx de Gizeh

                      Image
                      Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
                      Read more

                      Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

                      Image
                      3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
                      Read more