How do I use logarithmic differentiation to find the derivative of $y=x^{sin x}$?











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This is the work I've done so far:



$ln y=(sin x)ln$



$y'/y=(cos x)ln x+(sin x)/x$



And I'm not sure that I've set up the problem correctly...






calculus







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    Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 16:21






  • 1




    What steps did you get to derive what you have so far? Please share them in an edit.
    – Shaun
    Nov 26 at 16:21












  • There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
    – Lee Mosher
    Nov 26 at 16:56

















up vote
0
down vote

favorite












This is the work I've done so far:



$ln y=(sin x)ln$



$y'/y=(cos x)ln x+(sin x)/x$



And I'm not sure that I've set up the problem correctly...






calculus







share|cite|improve this question




















  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 16:21






  • 1




    What steps did you get to derive what you have so far? Please share them in an edit.
    – Shaun
    Nov 26 at 16:21












  • There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
    – Lee Mosher
    Nov 26 at 16:56















up vote
0
down vote

favorite









up vote
0
down vote

favorite











This is the work I've done so far:



$ln y=(sin x)ln$



$y'/y=(cos x)ln x+(sin x)/x$



And I'm not sure that I've set up the problem correctly...






calculus







share|cite|improve this question















This is the work I've done so far:



$ln y=(sin x)ln$



$y'/y=(cos x)ln x+(sin x)/x$



And I'm not sure that I've set up the problem correctly...






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 16:42









Larry

1,2712722




1,2712722










asked Nov 26 at 16:19









user597553

133




133








  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 16:21






  • 1




    What steps did you get to derive what you have so far? Please share them in an edit.
    – Shaun
    Nov 26 at 16:21












  • There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
    – Lee Mosher
    Nov 26 at 16:56
















  • 1




    Here's a MathJax tutorial :)
    – Shaun
    Nov 26 at 16:21






  • 1




    What steps did you get to derive what you have so far? Please share them in an edit.
    – Shaun
    Nov 26 at 16:21












  • There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
    – Lee Mosher
    Nov 26 at 16:56










1




1




Here's a MathJax tutorial :)
– Shaun
Nov 26 at 16:21




Here's a MathJax tutorial :)
– Shaun
Nov 26 at 16:21




1




1




What steps did you get to derive what you have so far? Please share them in an edit.
– Shaun
Nov 26 at 16:21






What steps did you get to derive what you have so far? Please share them in an edit.
– Shaun
Nov 26 at 16:21














There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
– Lee Mosher
Nov 26 at 16:56






There is one typographical(?) error in the work you've done so far: the equation $ln y=(sin x)ln$ is missing an "$x$" after the second "$ln$".
– Lee Mosher
Nov 26 at 16:56












2 Answers
2






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Let $y=x^{sin(x)}$. then $ln(y)=sin(x)ln(x)$. So: $$frac{1}{y}frac{mathrm{d}y}{mathrm{d}x}=cos(x)ln(x)+frac{sin(x)}{x}$$ Then multiplying up and substituting $y=x^{sin(x)}$ we get: $$frac{mathrm{d}y}{mathrm{d}x}=x^{sin(x)}(cos(x)ln(x)+frac{sin(x)}{x})$$






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    You have done! Complete and find:
    $$
    y'=yleft[(cos x)ln(x)+frac{sin x}x right]=x^{sin x}left[(cos x)ln(x)+frac{sin x}x right]
    $$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      Let $y=x^{sin(x)}$. then $ln(y)=sin(x)ln(x)$. So: $$frac{1}{y}frac{mathrm{d}y}{mathrm{d}x}=cos(x)ln(x)+frac{sin(x)}{x}$$ Then multiplying up and substituting $y=x^{sin(x)}$ we get: $$frac{mathrm{d}y}{mathrm{d}x}=x^{sin(x)}(cos(x)ln(x)+frac{sin(x)}{x})$$






      share|cite|improve this answer

























        up vote
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        Let $y=x^{sin(x)}$. then $ln(y)=sin(x)ln(x)$. So: $$frac{1}{y}frac{mathrm{d}y}{mathrm{d}x}=cos(x)ln(x)+frac{sin(x)}{x}$$ Then multiplying up and substituting $y=x^{sin(x)}$ we get: $$frac{mathrm{d}y}{mathrm{d}x}=x^{sin(x)}(cos(x)ln(x)+frac{sin(x)}{x})$$






        share|cite|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          Let $y=x^{sin(x)}$. then $ln(y)=sin(x)ln(x)$. So: $$frac{1}{y}frac{mathrm{d}y}{mathrm{d}x}=cos(x)ln(x)+frac{sin(x)}{x}$$ Then multiplying up and substituting $y=x^{sin(x)}$ we get: $$frac{mathrm{d}y}{mathrm{d}x}=x^{sin(x)}(cos(x)ln(x)+frac{sin(x)}{x})$$






          share|cite|improve this answer












          Let $y=x^{sin(x)}$. then $ln(y)=sin(x)ln(x)$. So: $$frac{1}{y}frac{mathrm{d}y}{mathrm{d}x}=cos(x)ln(x)+frac{sin(x)}{x}$$ Then multiplying up and substituting $y=x^{sin(x)}$ we get: $$frac{mathrm{d}y}{mathrm{d}x}=x^{sin(x)}(cos(x)ln(x)+frac{sin(x)}{x})$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 26 at 16:29









          Alfie

          211




          211






















              up vote
              1
              down vote













              You have done! Complete and find:
              $$
              y'=yleft[(cos x)ln(x)+frac{sin x}x right]=x^{sin x}left[(cos x)ln(x)+frac{sin x}x right]
              $$






              share|cite|improve this answer

























                up vote
                1
                down vote













                You have done! Complete and find:
                $$
                y'=yleft[(cos x)ln(x)+frac{sin x}x right]=x^{sin x}left[(cos x)ln(x)+frac{sin x}x right]
                $$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  You have done! Complete and find:
                  $$
                  y'=yleft[(cos x)ln(x)+frac{sin x}x right]=x^{sin x}left[(cos x)ln(x)+frac{sin x}x right]
                  $$






                  share|cite|improve this answer












                  You have done! Complete and find:
                  $$
                  y'=yleft[(cos x)ln(x)+frac{sin x}x right]=x^{sin x}left[(cos x)ln(x)+frac{sin x}x right]
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 26 at 16:24









                  Emilio Novati

                  51.2k43472




                  51.2k43472






























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