Derivative by definition of $ f(x)= 2xsin(2x)$












1












$begingroup$


I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}



From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?






derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
    $endgroup$
    – TonyK
    Jan 18 at 13:37










  • $begingroup$
    Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
    $endgroup$
    – Jakcjones
    Jan 18 at 13:49












  • $begingroup$
    Yes...${}{}{}{}{}$
    $endgroup$
    – TonyK
    Jan 18 at 13:56










  • $begingroup$
    Why is what I did wrong, then? Can you help me understand?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:39










  • $begingroup$
    I mean why can't it be $2(sinxcosh + cosxsinh)?$
    $endgroup$
    – Jakcjones
    Jan 18 at 14:42
















1












$begingroup$


I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}



From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?






derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    $sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
    $endgroup$
    – TonyK
    Jan 18 at 13:37










  • $begingroup$
    Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
    $endgroup$
    – Jakcjones
    Jan 18 at 13:49












  • $begingroup$
    Yes...${}{}{}{}{}$
    $endgroup$
    – TonyK
    Jan 18 at 13:56










  • $begingroup$
    Why is what I did wrong, then? Can you help me understand?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:39










  • $begingroup$
    I mean why can't it be $2(sinxcosh + cosxsinh)?$
    $endgroup$
    – Jakcjones
    Jan 18 at 14:42














1












1








1





$begingroup$


I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}



From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?






derivatives







share|cite|improve this question











$endgroup$




I want to find the derivative of $f(x)=2xsin(2x)$ but I'm having trouble developing the trigonometric identities. Here's what I worked out:
begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2(x+h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{2(x+h)(2sin(x)cos(h)+cos(x)sin(h))-2xsin(2x)}{h}\
&=lim_{hto0}frac{4xsin(x)cos(h)+4hsin(x)cos(h)+4xcos(x)sin(h)+4hcos(x)sin(h)-2xsin(2x)}{h}end{align}



From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?






derivatives




derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 15:24









Thomas Shelby

4,8082727




4,8082727










asked Jan 18 at 13:25









JakcjonesJakcjones

828




828












  • $begingroup$
    $sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
    $endgroup$
    – TonyK
    Jan 18 at 13:37










  • $begingroup$
    Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
    $endgroup$
    – Jakcjones
    Jan 18 at 13:49












  • $begingroup$
    Yes...${}{}{}{}{}$
    $endgroup$
    – TonyK
    Jan 18 at 13:56










  • $begingroup$
    Why is what I did wrong, then? Can you help me understand?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:39










  • $begingroup$
    I mean why can't it be $2(sinxcosh + cosxsinh)?$
    $endgroup$
    – Jakcjones
    Jan 18 at 14:42


















  • $begingroup$
    $sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
    $endgroup$
    – TonyK
    Jan 18 at 13:37










  • $begingroup$
    Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
    $endgroup$
    – Jakcjones
    Jan 18 at 13:49












  • $begingroup$
    Yes...${}{}{}{}{}$
    $endgroup$
    – TonyK
    Jan 18 at 13:56










  • $begingroup$
    Why is what I did wrong, then? Can you help me understand?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:39










  • $begingroup$
    I mean why can't it be $2(sinxcosh + cosxsinh)?$
    $endgroup$
    – Jakcjones
    Jan 18 at 14:42
















$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37




$begingroup$
$sin(2(x+h))$ is not equal to $2sin x cos h + cos x sin h$. It's not even equal to $2(sin x cos h + cos x sin h)$.
$endgroup$
– TonyK
Jan 18 at 13:37












$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49






$begingroup$
Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$?
$endgroup$
– Jakcjones
Jan 18 at 13:49














$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56




$begingroup$
Yes...${}{}{}{}{}$
$endgroup$
– TonyK
Jan 18 at 13:56












$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39




$begingroup$
Why is what I did wrong, then? Can you help me understand?
$endgroup$
– Jakcjones
Jan 18 at 14:39












$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42




$begingroup$
I mean why can't it be $2(sinxcosh + cosxsinh)?$
$endgroup$
– Jakcjones
Jan 18 at 14:42










2 Answers
2






active

oldest

votes


















2












$begingroup$

begin{align}
f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
&=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
&=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
&=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
&=4xcos2x+2sin2x
end{align}






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you explain how you come from the 3rd equality left side to the 4th?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:16












  • $begingroup$
    $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
    $endgroup$
    – Thomas Shelby
    Jan 18 at 14:25






  • 1




    $begingroup$
    Ok, got it. Wasn't visualizing it. Ty
    $endgroup$
    – Jakcjones
    Jan 18 at 14:27



















3












$begingroup$

I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:



$$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$



$$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$






share|cite|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    begin{align}
    f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
    &=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
    &=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
    &=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
    &=4xcos2x+2sin2x
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can you explain how you come from the 3rd equality left side to the 4th?
      $endgroup$
      – Jakcjones
      Jan 18 at 14:16












    • $begingroup$
      $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
      $endgroup$
      – Thomas Shelby
      Jan 18 at 14:25






    • 1




      $begingroup$
      Ok, got it. Wasn't visualizing it. Ty
      $endgroup$
      – Jakcjones
      Jan 18 at 14:27
















    2












    $begingroup$

    begin{align}
    f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
    &=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
    &=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
    &=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
    &=4xcos2x+2sin2x
    end{align}






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can you explain how you come from the 3rd equality left side to the 4th?
      $endgroup$
      – Jakcjones
      Jan 18 at 14:16












    • $begingroup$
      $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
      $endgroup$
      – Thomas Shelby
      Jan 18 at 14:25






    • 1




      $begingroup$
      Ok, got it. Wasn't visualizing it. Ty
      $endgroup$
      – Jakcjones
      Jan 18 at 14:27














    2












    2








    2





    $begingroup$

    begin{align}
    f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
    &=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
    &=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
    &=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
    &=4xcos2x+2sin2x
    end{align}






    share|cite|improve this answer











    $endgroup$



    begin{align}
    f'(x)&=lim_{hto0}frac{2(x+h)sin(2x+2h)-2xsin2x}{h}\
    &=lim_{hto0}left(frac{2xsin(2x+2h)-2xsin2x}{h}+frac{2hsin(2x+2h)}{h}right)\
    &=lim_{hto0}frac{2x(sin(2x+2h)-sin2x)}{h}+lim_{hto0}frac{2hsin(2x+2h)}{h}\
    &=lim_{hto0}frac{2x(2sin(h)cos(2x+h))}{h}+lim_{hto0}frac{2sin(2x+2h)}{1}\
    &=4xcos2x+2sin2x
    end{align}







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 18 at 13:58

























    answered Jan 18 at 13:52









    Thomas ShelbyThomas Shelby

    4,8082727




    4,8082727












    • $begingroup$
      Can you explain how you come from the 3rd equality left side to the 4th?
      $endgroup$
      – Jakcjones
      Jan 18 at 14:16












    • $begingroup$
      $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
      $endgroup$
      – Thomas Shelby
      Jan 18 at 14:25






    • 1




      $begingroup$
      Ok, got it. Wasn't visualizing it. Ty
      $endgroup$
      – Jakcjones
      Jan 18 at 14:27


















    • $begingroup$
      Can you explain how you come from the 3rd equality left side to the 4th?
      $endgroup$
      – Jakcjones
      Jan 18 at 14:16












    • $begingroup$
      $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
      $endgroup$
      – Thomas Shelby
      Jan 18 at 14:25






    • 1




      $begingroup$
      Ok, got it. Wasn't visualizing it. Ty
      $endgroup$
      – Jakcjones
      Jan 18 at 14:27
















    $begingroup$
    Can you explain how you come from the 3rd equality left side to the 4th?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:16






    $begingroup$
    Can you explain how you come from the 3rd equality left side to the 4th?
    $endgroup$
    – Jakcjones
    Jan 18 at 14:16














    $begingroup$
    $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
    $endgroup$
    – Thomas Shelby
    Jan 18 at 14:25




    $begingroup$
    $sin A-sin B=2sinleft(frac {A-B}{2}right)cosleft(frac {A+B}{2}right)$.
    $endgroup$
    – Thomas Shelby
    Jan 18 at 14:25




    1




    1




    $begingroup$
    Ok, got it. Wasn't visualizing it. Ty
    $endgroup$
    – Jakcjones
    Jan 18 at 14:27




    $begingroup$
    Ok, got it. Wasn't visualizing it. Ty
    $endgroup$
    – Jakcjones
    Jan 18 at 14:27











    3












    $begingroup$

    I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:



    $$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$



    $$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:



      $$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$



      $$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:



        $$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$



        $$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$






        share|cite|improve this answer











        $endgroup$



        I think you may be on the right track, yet the limit for the derivative definition you're using makes things specially cumbersome. What about the following?:



        $$frac{2xsin2 x-2x_0sin2 x_0}{x-x_0}=frac{2xsin 2x-2xsin 2x_0+2xsin2 x_0-2x_0sin x_0}{x-x_0}=$$



        $$=2xfrac{sin 2x-sin2 x_0}{x-x_0}+2sin 2x_0frac{x-x_0}{x-x_0}xrightarrow[xto x_0]{}4xcos2 x_0+2sin2 x_0$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 18 at 14:08

























        answered Jan 18 at 13:31









        DonAntonioDonAntonio

        180k1495233




        180k1495233






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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