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.






calculus functional-analysis







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.






    calculus functional-analysis







    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.






      calculus functional-analysis







      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




      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

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          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











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                    7 Answers
                    7






                    active

                    oldest

                    votes








                    7 Answers
                    7






                    active

                    oldest

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                    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


















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