Finding the tenth derivative












4












$begingroup$


I came across this Question where i have to find $$f^{(10)}$$ for the following function at $x = 0$
$$f(x) = e^xsin x$$



I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    I came across this Question where i have to find $$f^{(10)}$$ for the following function at $x = 0$
    $$f(x) = e^xsin x$$



    I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.










    share|cite|improve this question











    $endgroup$















      4












      4








      4





      $begingroup$


      I came across this Question where i have to find $$f^{(10)}$$ for the following function at $x = 0$
      $$f(x) = e^xsin x$$



      I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.










      share|cite|improve this question











      $endgroup$




      I came across this Question where i have to find $$f^{(10)}$$ for the following function at $x = 0$
      $$f(x) = e^xsin x$$



      I tried differentiating a few times to get a pattern but didn’t get one, can someone provide the solution.







      calculus functional-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Bernard

      119k639112




      119k639112










      asked 1 hour ago









      user601297user601297

      1797




      1797






















          7 Answers
          7






          active

          oldest

          votes


















          2












          $begingroup$

          Hint:



          $$f(x)=e^xsin x$$
          $$f'(x)=e^x(sin x +cos x)$$
          $$f''(x)=e^x(sin x+cos x)+e^x(cos x -sin x)=2e^x(cos x)$$
          $$f'''(x)=e^x(2cos x)-e^x(2sin x)=2e^x(cos x-sin x)$$
          $$f^{iv}(x)=2e^x(cos x-sin x)-2e^x(cos x+sin x)=-4e^x(sin x)=-4f(x)$$






          share|cite|improve this answer









          $endgroup$





















            8












            $begingroup$

            Hint:



            As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.






            share|cite|improve this answer









            $endgroup$









            • 1




              $begingroup$
              This is by far the most efficient solution.
              $endgroup$
              – Yves Daoust
              49 mins ago



















            1












            $begingroup$

            Use Leibniz' Rule for higher derivatives of a product:
            $$frac{d^n}{dx^n}(uv)
            =frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
            +binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$

            In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
            $$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
            &qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
            &=10-120+252-120+10cr
            &=32 .cr}$$






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              $$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$



              So



              $$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$





              If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:



              $$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$






              share|cite|improve this answer











              $endgroup$





















                0












                $begingroup$

                One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.



                Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.



                Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$






                share|cite|improve this answer









                $endgroup$





















                  0












                  $begingroup$

                  Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.






                  share|cite|improve this answer









                  $endgroup$





















                    -1












                    $begingroup$

                    I get $f^{10}(x)=32e^xcos x$.



                    Here's what I did:
                    $f'(x)=e^x(sin x+cos x)implies f''(x)=2e^xcos ximplies f^3(x)=2e^x(cos x-sin x)implies f^4(x)=2e^x(-2sin x)=-4f(x)implies f^5(x)=-4f'(x)implies f^8(x)=-4f^4(x)=16f(x)implies f^{10}(x)=16f''(x)$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                      $endgroup$
                      – AlexanderJ93
                      42 mins ago










                    • $begingroup$
                      Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                      $endgroup$
                      – Rhys Hughes
                      39 mins ago










                    • $begingroup$
                      @RhysHughes &Alexander J93 thanks
                      $endgroup$
                      – Chris Custer
                      34 mins ago











                    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%2f3077641%2ffinding-the-tenth-derivative%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown

























                    7 Answers
                    7






                    active

                    oldest

                    votes








                    7 Answers
                    7






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes









                    2












                    $begingroup$

                    Hint:



                    $$f(x)=e^xsin x$$
                    $$f'(x)=e^x(sin x +cos x)$$
                    $$f''(x)=e^x(sin x+cos x)+e^x(cos x -sin x)=2e^x(cos x)$$
                    $$f'''(x)=e^x(2cos x)-e^x(2sin x)=2e^x(cos x-sin x)$$
                    $$f^{iv}(x)=2e^x(cos x-sin x)-2e^x(cos x+sin x)=-4e^x(sin x)=-4f(x)$$






                    share|cite|improve this answer









                    $endgroup$


















                      2












                      $begingroup$

                      Hint:



                      $$f(x)=e^xsin x$$
                      $$f'(x)=e^x(sin x +cos x)$$
                      $$f''(x)=e^x(sin x+cos x)+e^x(cos x -sin x)=2e^x(cos x)$$
                      $$f'''(x)=e^x(2cos x)-e^x(2sin x)=2e^x(cos x-sin x)$$
                      $$f^{iv}(x)=2e^x(cos x-sin x)-2e^x(cos x+sin x)=-4e^x(sin x)=-4f(x)$$






                      share|cite|improve this answer









                      $endgroup$
















                        2












                        2








                        2





                        $begingroup$

                        Hint:



                        $$f(x)=e^xsin x$$
                        $$f'(x)=e^x(sin x +cos x)$$
                        $$f''(x)=e^x(sin x+cos x)+e^x(cos x -sin x)=2e^x(cos x)$$
                        $$f'''(x)=e^x(2cos x)-e^x(2sin x)=2e^x(cos x-sin x)$$
                        $$f^{iv}(x)=2e^x(cos x-sin x)-2e^x(cos x+sin x)=-4e^x(sin x)=-4f(x)$$






                        share|cite|improve this answer









                        $endgroup$



                        Hint:



                        $$f(x)=e^xsin x$$
                        $$f'(x)=e^x(sin x +cos x)$$
                        $$f''(x)=e^x(sin x+cos x)+e^x(cos x -sin x)=2e^x(cos x)$$
                        $$f'''(x)=e^x(2cos x)-e^x(2sin x)=2e^x(cos x-sin x)$$
                        $$f^{iv}(x)=2e^x(cos x-sin x)-2e^x(cos x+sin x)=-4e^x(sin x)=-4f(x)$$







                        share|cite|improve this answer












                        share|cite|improve this answer



                        share|cite|improve this answer










                        answered 1 hour ago









                        Rhys HughesRhys Hughes

                        5,2591427




                        5,2591427























                            8












                            $begingroup$

                            Hint:



                            As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              This is by far the most efficient solution.
                              $endgroup$
                              – Yves Daoust
                              49 mins ago
















                            8












                            $begingroup$

                            Hint:



                            As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.






                            share|cite|improve this answer









                            $endgroup$









                            • 1




                              $begingroup$
                              This is by far the most efficient solution.
                              $endgroup$
                              – Yves Daoust
                              49 mins ago














                            8












                            8








                            8





                            $begingroup$

                            Hint:



                            As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            As $;mathrm e^xsin x=operatorname{Im}bigl(mathrm e^{(1+i)x}bigr)$, you have to find first the real and imaginary parts of $(1+i)^{10}$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            BernardBernard

                            119k639112




                            119k639112








                            • 1




                              $begingroup$
                              This is by far the most efficient solution.
                              $endgroup$
                              – Yves Daoust
                              49 mins ago














                            • 1




                              $begingroup$
                              This is by far the most efficient solution.
                              $endgroup$
                              – Yves Daoust
                              49 mins ago








                            1




                            1




                            $begingroup$
                            This is by far the most efficient solution.
                            $endgroup$
                            – Yves Daoust
                            49 mins ago




                            $begingroup$
                            This is by far the most efficient solution.
                            $endgroup$
                            – Yves Daoust
                            49 mins ago











                            1












                            $begingroup$

                            Use Leibniz' Rule for higher derivatives of a product:
                            $$frac{d^n}{dx^n}(uv)
                            =frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
                            +binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$

                            In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
                            $$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
                            &qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
                            &=10-120+252-120+10cr
                            &=32 .cr}$$






                            share|cite|improve this answer









                            $endgroup$


















                              1












                              $begingroup$

                              Use Leibniz' Rule for higher derivatives of a product:
                              $$frac{d^n}{dx^n}(uv)
                              =frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
                              +binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$

                              In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
                              $$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
                              &qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
                              &=10-120+252-120+10cr
                              &=32 .cr}$$






                              share|cite|improve this answer









                              $endgroup$
















                                1












                                1








                                1





                                $begingroup$

                                Use Leibniz' Rule for higher derivatives of a product:
                                $$frac{d^n}{dx^n}(uv)
                                =frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
                                +binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$

                                In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
                                $$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
                                &qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
                                &=10-120+252-120+10cr
                                &=32 .cr}$$






                                share|cite|improve this answer









                                $endgroup$



                                Use Leibniz' Rule for higher derivatives of a product:
                                $$frac{d^n}{dx^n}(uv)
                                =frac{d^nu}{dx^n}v+binom n1frac{d^{n-1}u}{dx^{n-1}}frac{dv}{dx}
                                +binom n2frac{d^{n-2}u}{dx^{n-2}}frac{d^2v}{dx^2}+cdots+ufrac{d^nv}{dx^n} .$$

                                In your case take $u=e^x$ and $v=sin x$. Since you are going to substitute $x=0$ after differentiating, all the $e^x$ terms will be $1$, all the $sin x$ terms will be $0$ and all the cos $x$ terms will be $1$ (though some of them will pick up a negative sign when you differentiate). So the answer is
                                $$eqalign{0+binom{10}11&{}+binom{10}20+binom{10}3(-1)+binom{10}40+binom{10}51cr
                                &qquad{}+binom{10}60+binom{10}7(-1)+binom{10}80+binom{10}91+binom{10}{10}0cr
                                &=10-120+252-120+10cr
                                &=32 .cr}$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 59 mins ago









                                DavidDavid

                                68k664126




                                68k664126























                                    1












                                    $begingroup$

                                    $$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$



                                    So



                                    $$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$





                                    If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:



                                    $$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$






                                    share|cite|improve this answer











                                    $endgroup$


















                                      1












                                      $begingroup$

                                      $$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$



                                      So



                                      $$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$





                                      If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:



                                      $$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$






                                      share|cite|improve this answer











                                      $endgroup$
















                                        1












                                        1








                                        1





                                        $begingroup$

                                        $$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$



                                        So



                                        $$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$





                                        If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:



                                        $$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$






                                        share|cite|improve this answer











                                        $endgroup$



                                        $$(e^x(acos x+bsin x))'=e^x(acos x+bsin x)+e^x(bcos x-asin x)=e^x((a+b)cos x+(b-a)sin x).$$



                                        So



                                        $$(0,1)to(1,1)to(2,0)to(2,-2)to(0,-4)to(-4,-4)to(-8,0)to(-8,8)to(0,16)to(16,16)to(32,0).$$





                                        If you divide by increasing powers of $2$, in pairs, the pattern emerges, with period $8$:



                                        $$(0,1)to(1,1)to(1,0)to(1,-1)to(0,-1)to(-1,-1)to(-1,0)to(-1,1)to(0,1)to(1,1)to(1,0).$$







                                        share|cite|improve this answer














                                        share|cite|improve this answer



                                        share|cite|improve this answer








                                        edited 50 mins ago

























                                        answered 55 mins ago









                                        Yves DaoustYves Daoust

                                        125k671222




                                        125k671222























                                            0












                                            $begingroup$

                                            One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.



                                            Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.



                                            Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$






                                            share|cite|improve this answer









                                            $endgroup$


















                                              0












                                              $begingroup$

                                              One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.



                                              Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.



                                              Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$






                                              share|cite|improve this answer









                                              $endgroup$
















                                                0












                                                0








                                                0





                                                $begingroup$

                                                One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.



                                                Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.



                                                Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$






                                                share|cite|improve this answer









                                                $endgroup$



                                                One trick here is to use $e^{ix}=cos x+isin x$ and define $g(x)=e^xcos x$ then $f(x)$ and $g(x)$ are both real functions.



                                                Let $h(x)=g(x)+if(x)=e^{(1+i)x}$ then the tenth derivative of $h(x)$ is $(1+i)^{10}h(x)$ and the tenth derivative of $f(x)$ is the imaginary part of this.



                                                Because you only want the value at $x=0$ you can evaluate there, with $h(0)=g(0)+if(0)$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 1 hour ago









                                                Mark BennetMark Bennet

                                                80.8k981179




                                                80.8k981179























                                                    0












                                                    $begingroup$

                                                    Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.






                                                    share|cite|improve this answer









                                                    $endgroup$


















                                                      0












                                                      $begingroup$

                                                      Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.






                                                      share|cite|improve this answer









                                                      $endgroup$
















                                                        0












                                                        0








                                                        0





                                                        $begingroup$

                                                        Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.






                                                        share|cite|improve this answer









                                                        $endgroup$



                                                        Use the formula $(fg)^{(n)}= sumlimits_{k=0}^{n} binom {n} {k} (f)^{(k)}(g)^{(n-k)}$.







                                                        share|cite|improve this answer












                                                        share|cite|improve this answer



                                                        share|cite|improve this answer










                                                        answered 1 hour ago









                                                        Kavi Rama MurthyKavi Rama Murthy

                                                        53.9k32055




                                                        53.9k32055























                                                            -1












                                                            $begingroup$

                                                            I get $f^{10}(x)=32e^xcos x$.



                                                            Here's what I did:
                                                            $f'(x)=e^x(sin x+cos x)implies f''(x)=2e^xcos ximplies f^3(x)=2e^x(cos x-sin x)implies f^4(x)=2e^x(-2sin x)=-4f(x)implies f^5(x)=-4f'(x)implies f^8(x)=-4f^4(x)=16f(x)implies f^{10}(x)=16f''(x)$.






                                                            share|cite|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                              $endgroup$
                                                              – AlexanderJ93
                                                              42 mins ago










                                                            • $begingroup$
                                                              Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                              $endgroup$
                                                              – Rhys Hughes
                                                              39 mins ago










                                                            • $begingroup$
                                                              @RhysHughes &Alexander J93 thanks
                                                              $endgroup$
                                                              – Chris Custer
                                                              34 mins ago
















                                                            -1












                                                            $begingroup$

                                                            I get $f^{10}(x)=32e^xcos x$.



                                                            Here's what I did:
                                                            $f'(x)=e^x(sin x+cos x)implies f''(x)=2e^xcos ximplies f^3(x)=2e^x(cos x-sin x)implies f^4(x)=2e^x(-2sin x)=-4f(x)implies f^5(x)=-4f'(x)implies f^8(x)=-4f^4(x)=16f(x)implies f^{10}(x)=16f''(x)$.






                                                            share|cite|improve this answer











                                                            $endgroup$













                                                            • $begingroup$
                                                              Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                              $endgroup$
                                                              – AlexanderJ93
                                                              42 mins ago










                                                            • $begingroup$
                                                              Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                              $endgroup$
                                                              – Rhys Hughes
                                                              39 mins ago










                                                            • $begingroup$
                                                              @RhysHughes &Alexander J93 thanks
                                                              $endgroup$
                                                              – Chris Custer
                                                              34 mins ago














                                                            -1












                                                            -1








                                                            -1





                                                            $begingroup$

                                                            I get $f^{10}(x)=32e^xcos x$.



                                                            Here's what I did:
                                                            $f'(x)=e^x(sin x+cos x)implies f''(x)=2e^xcos ximplies f^3(x)=2e^x(cos x-sin x)implies f^4(x)=2e^x(-2sin x)=-4f(x)implies f^5(x)=-4f'(x)implies f^8(x)=-4f^4(x)=16f(x)implies f^{10}(x)=16f''(x)$.






                                                            share|cite|improve this answer











                                                            $endgroup$



                                                            I get $f^{10}(x)=32e^xcos x$.



                                                            Here's what I did:
                                                            $f'(x)=e^x(sin x+cos x)implies f''(x)=2e^xcos ximplies f^3(x)=2e^x(cos x-sin x)implies f^4(x)=2e^x(-2sin x)=-4f(x)implies f^5(x)=-4f'(x)implies f^8(x)=-4f^4(x)=16f(x)implies f^{10}(x)=16f''(x)$.







                                                            share|cite|improve this answer














                                                            share|cite|improve this answer



                                                            share|cite|improve this answer








                                                            edited 36 mins ago

























                                                            answered 43 mins ago









                                                            Chris CusterChris Custer

                                                            11.2k3824




                                                            11.2k3824












                                                            • $begingroup$
                                                              Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                              $endgroup$
                                                              – AlexanderJ93
                                                              42 mins ago










                                                            • $begingroup$
                                                              Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                              $endgroup$
                                                              – Rhys Hughes
                                                              39 mins ago










                                                            • $begingroup$
                                                              @RhysHughes &Alexander J93 thanks
                                                              $endgroup$
                                                              – Chris Custer
                                                              34 mins ago


















                                                            • $begingroup$
                                                              Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                              $endgroup$
                                                              – AlexanderJ93
                                                              42 mins ago










                                                            • $begingroup$
                                                              Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                              $endgroup$
                                                              – Rhys Hughes
                                                              39 mins ago










                                                            • $begingroup$
                                                              @RhysHughes &Alexander J93 thanks
                                                              $endgroup$
                                                              – Chris Custer
                                                              34 mins ago
















                                                            $begingroup$
                                                            Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                            $endgroup$
                                                            – AlexanderJ93
                                                            42 mins ago




                                                            $begingroup$
                                                            Your calculation for $f^{(4)}$ is wrong. Looks like you missed a product rule.
                                                            $endgroup$
                                                            – AlexanderJ93
                                                            42 mins ago












                                                            $begingroup$
                                                            Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                            $endgroup$
                                                            – Rhys Hughes
                                                            39 mins ago




                                                            $begingroup$
                                                            Be careful: $$(e^x(g(x))'=e^xg(x)+e^xg'(x)$$ You missed the first bit in your calculation for $f^4$
                                                            $endgroup$
                                                            – Rhys Hughes
                                                            39 mins ago












                                                            $begingroup$
                                                            @RhysHughes &Alexander J93 thanks
                                                            $endgroup$
                                                            – Chris Custer
                                                            34 mins ago




                                                            $begingroup$
                                                            @RhysHughes &Alexander J93 thanks
                                                            $endgroup$
                                                            – Chris Custer
                                                            34 mins ago


















                                                            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.




                                                            draft saved


                                                            draft discarded














                                                            StackExchange.ready(
                                                            function () {
                                                            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3077641%2ffinding-the-tenth-derivative%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

                                                            Magento 2 controller redirect on button click in phtml file

                                                            Polycentropodidae