How to find the derivative of $cos 5x$ using first principle












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$begingroup$


So my textbook solution uses some discrete method to arrive at -5sinx 5x. But I want to know a simpler technique to get the solution. Can anyone post?






derivatives implicit-differentiation







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  • $begingroup$
    It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
    $endgroup$
    – String
    Jan 6 at 13:13
















0












$begingroup$


So my textbook solution uses some discrete method to arrive at -5sinx 5x. But I want to know a simpler technique to get the solution. Can anyone post?






derivatives implicit-differentiation







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
    $endgroup$
    – String
    Jan 6 at 13:13














0












0








0





$begingroup$


So my textbook solution uses some discrete method to arrive at -5sinx 5x. But I want to know a simpler technique to get the solution. Can anyone post?






derivatives implicit-differentiation







share|cite|improve this question









$endgroup$




So my textbook solution uses some discrete method to arrive at -5sinx 5x. But I want to know a simpler technique to get the solution. Can anyone post?






derivatives implicit-differentiation




derivatives implicit-differentiation






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asked Jan 6 at 12:46









Shaikh SakibShaikh Sakib

195




195












  • $begingroup$
    It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
    $endgroup$
    – String
    Jan 6 at 13:13


















  • $begingroup$
    It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
    $endgroup$
    – String
    Jan 6 at 13:13
















$begingroup$
It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
$endgroup$
– String
Jan 6 at 13:13




$begingroup$
It is rather hard to tell what you are actually after. Do you want an in depth formal proof or just a different pre-packed technique? Also we do not know what method your textbook is using. Can you clarify?
$endgroup$
– String
Jan 6 at 13:13










2 Answers
2






active

oldest

votes


















1












$begingroup$

By the definition of a derivative, you have



$$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h}$$



So using $f(x) = cos(5x)$, you get



$$lim_{h to 0}frac{cos(5x+h)-cos(5x)}{h}$$



To get you started, you need to know:




  • $$cos(alpha+beta) = cos (alpha)cos (beta)-sin (alpha)sin (beta)$$

  • $$color{blue}{lim_{x to 0} frac{sin(x)}{x} = 1}$$

  • $$lim_{x to 0}frac{cos(x)-1}{x} = 0$$


For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $lim_limits{x to 0}frac{sin(nx)}{x}$. Can you take it on from here?






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $$lim_{hrightarrow0}dfrac{f(x+h)-f(x)}{h}$$
    $$implies lim_{h to 0}frac{cos5(x+h)-cos5(x)}{h}$$
    $$implies lim_{h to 0} frac{left(1-frac{1}{2}5^2(x+h)^2+frac{1}{24}5^4(x+h)^4-.... right)-left((1-frac{1}{2}5^2(x)^2+frac{1}{24}5^4(x)^4-.... right)}{h}$$



    $$implies lim_{h to 0}frac{-frac{1}{2}5^2(2xh+h^2)+frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$
    $$implies lim_{h to 0} frac{hleft[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- right]}{h}$$
    $$implies lim_{h to 0}left[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... right]$$
    $$implies -5left[frac{1}{2}5(2x)-frac{1}{24}5^3(4x^3)+... right] $$
    $$implies -5left[(5x)-frac{1}{6}(5x)^3+... right] $$
    $$implies -5sin 5x$$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      1












      $begingroup$

      By the definition of a derivative, you have



      $$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h}$$



      So using $f(x) = cos(5x)$, you get



      $$lim_{h to 0}frac{cos(5x+h)-cos(5x)}{h}$$



      To get you started, you need to know:




      • $$cos(alpha+beta) = cos (alpha)cos (beta)-sin (alpha)sin (beta)$$

      • $$color{blue}{lim_{x to 0} frac{sin(x)}{x} = 1}$$

      • $$lim_{x to 0}frac{cos(x)-1}{x} = 0$$


      For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $lim_limits{x to 0}frac{sin(nx)}{x}$. Can you take it on from here?






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        By the definition of a derivative, you have



        $$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h}$$



        So using $f(x) = cos(5x)$, you get



        $$lim_{h to 0}frac{cos(5x+h)-cos(5x)}{h}$$



        To get you started, you need to know:




        • $$cos(alpha+beta) = cos (alpha)cos (beta)-sin (alpha)sin (beta)$$

        • $$color{blue}{lim_{x to 0} frac{sin(x)}{x} = 1}$$

        • $$lim_{x to 0}frac{cos(x)-1}{x} = 0$$


        For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $lim_limits{x to 0}frac{sin(nx)}{x}$. Can you take it on from here?






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          By the definition of a derivative, you have



          $$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h}$$



          So using $f(x) = cos(5x)$, you get



          $$lim_{h to 0}frac{cos(5x+h)-cos(5x)}{h}$$



          To get you started, you need to know:




          • $$cos(alpha+beta) = cos (alpha)cos (beta)-sin (alpha)sin (beta)$$

          • $$color{blue}{lim_{x to 0} frac{sin(x)}{x} = 1}$$

          • $$lim_{x to 0}frac{cos(x)-1}{x} = 0$$


          For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $lim_limits{x to 0}frac{sin(nx)}{x}$. Can you take it on from here?






          share|cite|improve this answer









          $endgroup$



          By the definition of a derivative, you have



          $$f’(x) = lim_{h to 0}frac{f(x+h)-f(x)}{h}$$



          So using $f(x) = cos(5x)$, you get



          $$lim_{h to 0}frac{cos(5x+h)-cos(5x)}{h}$$



          To get you started, you need to know:




          • $$cos(alpha+beta) = cos (alpha)cos (beta)-sin (alpha)sin (beta)$$

          • $$color{blue}{lim_{x to 0} frac{sin(x)}{x} = 1}$$

          • $$lim_{x to 0}frac{cos(x)-1}{x} = 0$$


          For the limit highlighted in blue, you need a simple manipulation since you get a limit in the form $lim_limits{x to 0}frac{sin(nx)}{x}$. Can you take it on from here?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 12:56









          KM101KM101

          5,9851524




          5,9851524























              0












              $begingroup$

              $$lim_{hrightarrow0}dfrac{f(x+h)-f(x)}{h}$$
              $$implies lim_{h to 0}frac{cos5(x+h)-cos5(x)}{h}$$
              $$implies lim_{h to 0} frac{left(1-frac{1}{2}5^2(x+h)^2+frac{1}{24}5^4(x+h)^4-.... right)-left((1-frac{1}{2}5^2(x)^2+frac{1}{24}5^4(x)^4-.... right)}{h}$$



              $$implies lim_{h to 0}frac{-frac{1}{2}5^2(2xh+h^2)+frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$
              $$implies lim_{h to 0} frac{hleft[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- right]}{h}$$
              $$implies lim_{h to 0}left[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... right]$$
              $$implies -5left[frac{1}{2}5(2x)-frac{1}{24}5^3(4x^3)+... right] $$
              $$implies -5left[(5x)-frac{1}{6}(5x)^3+... right] $$
              $$implies -5sin 5x$$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$lim_{hrightarrow0}dfrac{f(x+h)-f(x)}{h}$$
                $$implies lim_{h to 0}frac{cos5(x+h)-cos5(x)}{h}$$
                $$implies lim_{h to 0} frac{left(1-frac{1}{2}5^2(x+h)^2+frac{1}{24}5^4(x+h)^4-.... right)-left((1-frac{1}{2}5^2(x)^2+frac{1}{24}5^4(x)^4-.... right)}{h}$$



                $$implies lim_{h to 0}frac{-frac{1}{2}5^2(2xh+h^2)+frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$
                $$implies lim_{h to 0} frac{hleft[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- right]}{h}$$
                $$implies lim_{h to 0}left[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... right]$$
                $$implies -5left[frac{1}{2}5(2x)-frac{1}{24}5^3(4x^3)+... right] $$
                $$implies -5left[(5x)-frac{1}{6}(5x)^3+... right] $$
                $$implies -5sin 5x$$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$lim_{hrightarrow0}dfrac{f(x+h)-f(x)}{h}$$
                  $$implies lim_{h to 0}frac{cos5(x+h)-cos5(x)}{h}$$
                  $$implies lim_{h to 0} frac{left(1-frac{1}{2}5^2(x+h)^2+frac{1}{24}5^4(x+h)^4-.... right)-left((1-frac{1}{2}5^2(x)^2+frac{1}{24}5^4(x)^4-.... right)}{h}$$



                  $$implies lim_{h to 0}frac{-frac{1}{2}5^2(2xh+h^2)+frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$
                  $$implies lim_{h to 0} frac{hleft[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- right]}{h}$$
                  $$implies lim_{h to 0}left[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... right]$$
                  $$implies -5left[frac{1}{2}5(2x)-frac{1}{24}5^3(4x^3)+... right] $$
                  $$implies -5left[(5x)-frac{1}{6}(5x)^3+... right] $$
                  $$implies -5sin 5x$$






                  share|cite|improve this answer









                  $endgroup$



                  $$lim_{hrightarrow0}dfrac{f(x+h)-f(x)}{h}$$
                  $$implies lim_{h to 0}frac{cos5(x+h)-cos5(x)}{h}$$
                  $$implies lim_{h to 0} frac{left(1-frac{1}{2}5^2(x+h)^2+frac{1}{24}5^4(x+h)^4-.... right)-left((1-frac{1}{2}5^2(x)^2+frac{1}{24}5^4(x)^4-.... right)}{h}$$



                  $$implies lim_{h to 0}frac{-frac{1}{2}5^2(2xh+h^2)+frac{1}{24}5^4(4x^3h+6x^2h^2+4xh^3+h^4)...-}{h}$$
                  $$implies lim_{h to 0} frac{hleft[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)...- right]}{h}$$
                  $$implies lim_{h to 0}left[-frac{1}{2}5^2(2x+h)+frac{1}{24}5^4(4x^3+6x^2h+4xh^2+h^3)-... right]$$
                  $$implies -5left[frac{1}{2}5(2x)-frac{1}{24}5^3(4x^3)+... right] $$
                  $$implies -5left[(5x)-frac{1}{6}(5x)^3+... right] $$
                  $$implies -5sin 5x$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 13:22









                  Rakibul Islam PrinceRakibul Islam Prince

                  988211




                  988211






























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