Why does the derivative of $sin^2x$ need the chain rule? Isn't it just $2sin x$ by the power rule?











up vote
2
down vote

favorite












It's probably something obvious and I'm gonna slap myself in the face again, but




Why is the first derivative of $sin^2(x)$ calculated via the chain rule? Isn't it just a standard $x^a$ (power rule) case, and therefore just $2sin(x)$?











share|cite|improve this question




















  • 1




    $x$ isn’t $sin x$.
    – Randall
    Nov 17 at 14:26






  • 3




    No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
    – user3482749
    Nov 17 at 14:27















up vote
2
down vote

favorite












It's probably something obvious and I'm gonna slap myself in the face again, but




Why is the first derivative of $sin^2(x)$ calculated via the chain rule? Isn't it just a standard $x^a$ (power rule) case, and therefore just $2sin(x)$?











share|cite|improve this question




















  • 1




    $x$ isn’t $sin x$.
    – Randall
    Nov 17 at 14:26






  • 3




    No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
    – user3482749
    Nov 17 at 14:27













up vote
2
down vote

favorite









up vote
2
down vote

favorite











It's probably something obvious and I'm gonna slap myself in the face again, but




Why is the first derivative of $sin^2(x)$ calculated via the chain rule? Isn't it just a standard $x^a$ (power rule) case, and therefore just $2sin(x)$?











share|cite|improve this question















It's probably something obvious and I'm gonna slap myself in the face again, but




Why is the first derivative of $sin^2(x)$ calculated via the chain rule? Isn't it just a standard $x^a$ (power rule) case, and therefore just $2sin(x)$?








calculus functions trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 17 at 19:18









Blue

46.9k870147




46.9k870147










asked Nov 17 at 14:25









Arcturus

475




475








  • 1




    $x$ isn’t $sin x$.
    – Randall
    Nov 17 at 14:26






  • 3




    No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
    – user3482749
    Nov 17 at 14:27














  • 1




    $x$ isn’t $sin x$.
    – Randall
    Nov 17 at 14:26






  • 3




    No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
    – user3482749
    Nov 17 at 14:27








1




1




$x$ isn’t $sin x$.
– Randall
Nov 17 at 14:26




$x$ isn’t $sin x$.
– Randall
Nov 17 at 14:26




3




3




No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
– user3482749
Nov 17 at 14:27




No. There's quite clearly a "$sin$" in there. $sin^2(x) = (sin(x))^2$. Notice how we've composed two functions together? That means that we need the chain rule.
– user3482749
Nov 17 at 14:27










6 Answers
6






active

oldest

votes

















up vote
1
down vote













$sin xne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=sin x$.






share|cite|improve this answer





















  • Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
    – Arcturus
    Nov 17 at 14:33


















up vote
1
down vote













Maybe a more convincing example. Following your logics,



$$((x^3)^2)'=2x^3.$$



But don't we have



$$((x^3)^2)'=(x^6)'=6x^5 ?!$$






share|cite|improve this answer




























    up vote
    0
    down vote













    No it is a case



    $$g(x)=[f(x)]^2implies g’(x)=2f(x)f’(x)$$






    share|cite|improve this answer




























      up vote
      0
      down vote













      $sin^2 x$ means $(sin x)^2$. You’re asked to differentiate with respect to $x$, not $sin x$, so the Chain Rule is required. So, this is a case of a composite function.



      $$(f circ g)’(x) = f’(g(x))cdot g’$$



      In this case, you have



      $$big(u^2big)’ = 2ucdot u’$$



      where $u = sin x$.



      If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.






      share|cite|improve this answer




























        up vote
        0
        down vote













        No! More general you got for the function $f(x)=x^a$ we get as derivative



        $$f'(x)=(x^a)'=a(x)^{a-1}cdot(x)'=a(x)^{a-1}cdot1=ax^{a-1}$$



        This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))cdot g'(x)$. Therefore for $f(x)=sin^2(x)$ we got



        $$f'(x)=(sin^2(x))'=2sin(x)cdot(sin(x))'=2sin(x)cdotcos(x)=2sin(x)cos(x)$$



        where the inner function is given by $g(x)=sin(x)$ and the outer function $f(x)=x^2$.






        share|cite|improve this answer






























          up vote
          0
          down vote













          The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $sin^2(x)=sin(x)cdotsin(x)$. Now apply the product rule.






          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',
            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%2f3002408%2fwhy-does-the-derivative-of-sin2x-need-the-chain-rule-isnt-it-just-2-sin-x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            1
            down vote













            $sin xne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=sin x$.






            share|cite|improve this answer





















            • Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
              – Arcturus
              Nov 17 at 14:33















            up vote
            1
            down vote













            $sin xne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=sin x$.






            share|cite|improve this answer





















            • Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
              – Arcturus
              Nov 17 at 14:33













            up vote
            1
            down vote










            up vote
            1
            down vote









            $sin xne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=sin x$.






            share|cite|improve this answer












            $sin xne x$, so using $x^a$ alone is not enough. In the $f(g(x))$ of the chain rule, $f(x)=x^2$ and $g(x)=sin x$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 17 at 14:27









            Parcly Taxel

            41k137198




            41k137198












            • Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
              – Arcturus
              Nov 17 at 14:33


















            • Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
              – Arcturus
              Nov 17 at 14:33
















            Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
            – Arcturus
            Nov 17 at 14:33




            Right. I knew it was something obvious that I was missing. $f(x)$ was $x^2$. Thanks!
            – Arcturus
            Nov 17 at 14:33










            up vote
            1
            down vote













            Maybe a more convincing example. Following your logics,



            $$((x^3)^2)'=2x^3.$$



            But don't we have



            $$((x^3)^2)'=(x^6)'=6x^5 ?!$$






            share|cite|improve this answer

























              up vote
              1
              down vote













              Maybe a more convincing example. Following your logics,



              $$((x^3)^2)'=2x^3.$$



              But don't we have



              $$((x^3)^2)'=(x^6)'=6x^5 ?!$$






              share|cite|improve this answer























                up vote
                1
                down vote










                up vote
                1
                down vote









                Maybe a more convincing example. Following your logics,



                $$((x^3)^2)'=2x^3.$$



                But don't we have



                $$((x^3)^2)'=(x^6)'=6x^5 ?!$$






                share|cite|improve this answer












                Maybe a more convincing example. Following your logics,



                $$((x^3)^2)'=2x^3.$$



                But don't we have



                $$((x^3)^2)'=(x^6)'=6x^5 ?!$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 14:37









                Yves Daoust

                122k668217




                122k668217






















                    up vote
                    0
                    down vote













                    No it is a case



                    $$g(x)=[f(x)]^2implies g’(x)=2f(x)f’(x)$$






                    share|cite|improve this answer

























                      up vote
                      0
                      down vote













                      No it is a case



                      $$g(x)=[f(x)]^2implies g’(x)=2f(x)f’(x)$$






                      share|cite|improve this answer























                        up vote
                        0
                        down vote










                        up vote
                        0
                        down vote









                        No it is a case



                        $$g(x)=[f(x)]^2implies g’(x)=2f(x)f’(x)$$






                        share|cite|improve this answer












                        No it is a case



                        $$g(x)=[f(x)]^2implies g’(x)=2f(x)f’(x)$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered Nov 17 at 14:27









                        gimusi

                        88.5k74394




                        88.5k74394






















                            up vote
                            0
                            down vote













                            $sin^2 x$ means $(sin x)^2$. You’re asked to differentiate with respect to $x$, not $sin x$, so the Chain Rule is required. So, this is a case of a composite function.



                            $$(f circ g)’(x) = f’(g(x))cdot g’$$



                            In this case, you have



                            $$big(u^2big)’ = 2ucdot u’$$



                            where $u = sin x$.



                            If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.






                            share|cite|improve this answer

























                              up vote
                              0
                              down vote













                              $sin^2 x$ means $(sin x)^2$. You’re asked to differentiate with respect to $x$, not $sin x$, so the Chain Rule is required. So, this is a case of a composite function.



                              $$(f circ g)’(x) = f’(g(x))cdot g’$$



                              In this case, you have



                              $$big(u^2big)’ = 2ucdot u’$$



                              where $u = sin x$.



                              If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.






                              share|cite|improve this answer























                                up vote
                                0
                                down vote










                                up vote
                                0
                                down vote









                                $sin^2 x$ means $(sin x)^2$. You’re asked to differentiate with respect to $x$, not $sin x$, so the Chain Rule is required. So, this is a case of a composite function.



                                $$(f circ g)’(x) = f’(g(x))cdot g’$$



                                In this case, you have



                                $$big(u^2big)’ = 2ucdot u’$$



                                where $u = sin x$.



                                If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.






                                share|cite|improve this answer












                                $sin^2 x$ means $(sin x)^2$. You’re asked to differentiate with respect to $x$, not $sin x$, so the Chain Rule is required. So, this is a case of a composite function.



                                $$(f circ g)’(x) = f’(g(x))cdot g’$$



                                In this case, you have



                                $$big(u^2big)’ = 2ucdot u’$$



                                where $u = sin x$.



                                If, for whatever reason, you wanted to differentiate $u^2$ with respect to $u$, then you would have $2u$, but that isn’t the case.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Nov 17 at 14:28









                                KM101

                                2,727416




                                2,727416






















                                    up vote
                                    0
                                    down vote













                                    No! More general you got for the function $f(x)=x^a$ we get as derivative



                                    $$f'(x)=(x^a)'=a(x)^{a-1}cdot(x)'=a(x)^{a-1}cdot1=ax^{a-1}$$



                                    This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))cdot g'(x)$. Therefore for $f(x)=sin^2(x)$ we got



                                    $$f'(x)=(sin^2(x))'=2sin(x)cdot(sin(x))'=2sin(x)cdotcos(x)=2sin(x)cos(x)$$



                                    where the inner function is given by $g(x)=sin(x)$ and the outer function $f(x)=x^2$.






                                    share|cite|improve this answer



























                                      up vote
                                      0
                                      down vote













                                      No! More general you got for the function $f(x)=x^a$ we get as derivative



                                      $$f'(x)=(x^a)'=a(x)^{a-1}cdot(x)'=a(x)^{a-1}cdot1=ax^{a-1}$$



                                      This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))cdot g'(x)$. Therefore for $f(x)=sin^2(x)$ we got



                                      $$f'(x)=(sin^2(x))'=2sin(x)cdot(sin(x))'=2sin(x)cdotcos(x)=2sin(x)cos(x)$$



                                      where the inner function is given by $g(x)=sin(x)$ and the outer function $f(x)=x^2$.






                                      share|cite|improve this answer

























                                        up vote
                                        0
                                        down vote










                                        up vote
                                        0
                                        down vote









                                        No! More general you got for the function $f(x)=x^a$ we get as derivative



                                        $$f'(x)=(x^a)'=a(x)^{a-1}cdot(x)'=a(x)^{a-1}cdot1=ax^{a-1}$$



                                        This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))cdot g'(x)$. Therefore for $f(x)=sin^2(x)$ we got



                                        $$f'(x)=(sin^2(x))'=2sin(x)cdot(sin(x))'=2sin(x)cdotcos(x)=2sin(x)cos(x)$$



                                        where the inner function is given by $g(x)=sin(x)$ and the outer function $f(x)=x^2$.






                                        share|cite|improve this answer














                                        No! More general you got for the function $f(x)=x^a$ we get as derivative



                                        $$f'(x)=(x^a)'=a(x)^{a-1}cdot(x)'=a(x)^{a-1}cdot1=ax^{a-1}$$



                                        This only works out since the derivatives of the inner function, $x$, equals $1$. For the case that we got another function $g(x)$ as inner function we are forced to use the chain rule and thus $(f(g(x)))'=f'(g(x))cdot g'(x)$. Therefore for $f(x)=sin^2(x)$ we got



                                        $$f'(x)=(sin^2(x))'=2sin(x)cdot(sin(x))'=2sin(x)cdotcos(x)=2sin(x)cos(x)$$



                                        where the inner function is given by $g(x)=sin(x)$ and the outer function $f(x)=x^2$.







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited Nov 17 at 14:47

























                                        answered Nov 17 at 14:29









                                        mrtaurho

                                        2,5941827




                                        2,5941827






















                                            up vote
                                            0
                                            down vote













                                            The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $sin^2(x)=sin(x)cdotsin(x)$. Now apply the product rule.






                                            share|cite|improve this answer

























                                              up vote
                                              0
                                              down vote













                                              The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $sin^2(x)=sin(x)cdotsin(x)$. Now apply the product rule.






                                              share|cite|improve this answer























                                                up vote
                                                0
                                                down vote










                                                up vote
                                                0
                                                down vote









                                                The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $sin^2(x)=sin(x)cdotsin(x)$. Now apply the product rule.






                                                share|cite|improve this answer












                                                The power rule can only be applied to take the derivative of a power of $x$, which isn't the case here. Write $sin^2(x)=sin(x)cdotsin(x)$. Now apply the product rule.







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered Nov 17 at 18:32









                                                Michael Hoppe

                                                10.6k31733




                                                10.6k31733






























                                                    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%2f3002408%2fwhy-does-the-derivative-of-sin2x-need-the-chain-rule-isnt-it-just-2-sin-x%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

                                                    AnyDesk - Fatal Program Failure

                                                    How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                                                    QoS: MAC-Priority for clients behind a repeater