What is derivative of $sin ax$ where $a$ is a constant?












1














What is the derivative of $sin a x$ where $a$ is a constant.




Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.



I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.




Thank You!






calculus algebra-precalculus derivatives differential







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  • 1




    Apply the chain rule.
    – user3482749
    Nov 29 at 15:17
















1














What is the derivative of $sin a x$ where $a$ is a constant.




Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.



I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.




Thank You!






calculus algebra-precalculus derivatives differential







share|cite|improve this question




















  • 1




    Apply the chain rule.
    – user3482749
    Nov 29 at 15:17














1












1








1







What is the derivative of $sin a x$ where $a$ is a constant.




Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.



I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.




Thank You!






calculus algebra-precalculus derivatives differential







share|cite|improve this question















What is the derivative of $sin a x$ where $a$ is a constant.




Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.



I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.




Thank You!






calculus algebra-precalculus derivatives differential




calculus algebra-precalculus derivatives differential






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 at 19:19









Asaf Karagila♦

301k32423755




301k32423755










asked Nov 29 at 15:17









Chaku Daku

91




91








  • 1




    Apply the chain rule.
    – user3482749
    Nov 29 at 15:17














  • 1




    Apply the chain rule.
    – user3482749
    Nov 29 at 15:17








1




1




Apply the chain rule.
– user3482749
Nov 29 at 15:17




Apply the chain rule.
– user3482749
Nov 29 at 15:17










3 Answers
3






active

oldest

votes


















6














HINT



Recall that by chain rule



$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$






share|cite|improve this answer





























    3














    begin{array}{c}
    frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
    = left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
    = left( {cos ax} right) cdot a cdot 1\
    = aleft( {cos ax} right)
    end{array}






    share|cite|improve this answer

















    • 1




      Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
      – Chris
      Nov 29 at 17:20






    • 1




      @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
      – Paul Sinclair
      Nov 29 at 17:29



















    1














    Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      HINT



      Recall that by chain rule



      $$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$






      share|cite|improve this answer


























        6














        HINT



        Recall that by chain rule



        $$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$






        share|cite|improve this answer
























          6












          6








          6






          HINT



          Recall that by chain rule



          $$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$






          share|cite|improve this answer












          HINT



          Recall that by chain rule



          $$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 at 15:20









          gimusi

          1




          1























              3














              begin{array}{c}
              frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
              = left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
              = left( {cos ax} right) cdot a cdot 1\
              = aleft( {cos ax} right)
              end{array}






              share|cite|improve this answer

















              • 1




                Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
                – Chris
                Nov 29 at 17:20






              • 1




                @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
                – Paul Sinclair
                Nov 29 at 17:29
















              3














              begin{array}{c}
              frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
              = left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
              = left( {cos ax} right) cdot a cdot 1\
              = aleft( {cos ax} right)
              end{array}






              share|cite|improve this answer

















              • 1




                Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
                – Chris
                Nov 29 at 17:20






              • 1




                @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
                – Paul Sinclair
                Nov 29 at 17:29














              3












              3








              3






              begin{array}{c}
              frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
              = left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
              = left( {cos ax} right) cdot a cdot 1\
              = aleft( {cos ax} right)
              end{array}






              share|cite|improve this answer












              begin{array}{c}
              frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
              = left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
              = left( {cos ax} right) cdot a cdot 1\
              = aleft( {cos ax} right)
              end{array}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 29 at 16:10









              Krishna Srivastav

              894




              894








              • 1




                Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
                – Chris
                Nov 29 at 17:20






              • 1




                @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
                – Paul Sinclair
                Nov 29 at 17:29














              • 1




                Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
                – Chris
                Nov 29 at 17:20






              • 1




                @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
                – Paul Sinclair
                Nov 29 at 17:29








              1




              1




              Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
              – Chris
              Nov 29 at 17:20




              Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
              – Chris
              Nov 29 at 17:20




              1




              1




              @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
              – Paul Sinclair
              Nov 29 at 17:29




              @Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
              – Paul Sinclair
              Nov 29 at 17:29











              1














              Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.






              share|cite|improve this answer




























                1














                Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.






                share|cite|improve this answer


























                  1












                  1








                  1






                  Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.






                  share|cite|improve this answer














                  Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 at 15:42









                  Tianlalu

                  3,05521038




                  3,05521038










                  answered Nov 29 at 15:21









                  Rohit Bharadwaj

                  518




                  518






























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