Why is u'(t)/u(t) = (ln(u(t)))'?

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up vote
2
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I saw this from these online notes about differential equations.




$fracmu'(t)mu(t) = p(t)$



Hopefully you will recognize the left side of this from your Calculus I class as the following derivative.



($lnmu(t))' = p(t)$




I don't think I understand why $fracmu'(t)mu(t) = (lnmu(t))'$ is true. Why is this the case? Thank you very much!




calculus differential-equations




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  • 3




    Two words: chain rule.
    – Sean Roberson
    yesterday










  • Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
    – D. Beec
    yesterday











  • Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
    – Beneschan
    yesterday














up vote
2
down vote

favorite












I saw this from these online notes about differential equations.




$fracmu'(t)mu(t) = p(t)$



Hopefully you will recognize the left side of this from your Calculus I class as the following derivative.



($lnmu(t))' = p(t)$




I don't think I understand why $fracmu'(t)mu(t) = (lnmu(t))'$ is true. Why is this the case? Thank you very much!




calculus differential-equations




share|cite|improve this question















  • 3




    Two words: chain rule.
    – Sean Roberson
    yesterday










  • Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
    – D. Beec
    yesterday











  • Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
    – Beneschan
    yesterday












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I saw this from these online notes about differential equations.




$fracmu'(t)mu(t) = p(t)$



Hopefully you will recognize the left side of this from your Calculus I class as the following derivative.



($lnmu(t))' = p(t)$




I don't think I understand why $fracmu'(t)mu(t) = (lnmu(t))'$ is true. Why is this the case? Thank you very much!




calculus differential-equations




share|cite|improve this question











I saw this from these online notes about differential equations.




$fracmu'(t)mu(t) = p(t)$



Hopefully you will recognize the left side of this from your Calculus I class as the following derivative.



($lnmu(t))' = p(t)$




I don't think I understand why $fracmu'(t)mu(t) = (lnmu(t))'$ is true. Why is this the case? Thank you very much!





calculus differential-equations





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked yesterday









Beneschan

44537




44537







  • 3




    Two words: chain rule.
    – Sean Roberson
    yesterday










  • Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
    – D. Beec
    yesterday











  • Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
    – Beneschan
    yesterday












  • 3




    Two words: chain rule.
    – Sean Roberson
    yesterday










  • Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
    – D. Beec
    yesterday











  • Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
    – Beneschan
    yesterday







3




3




Two words: chain rule.
– Sean Roberson
yesterday




Two words: chain rule.
– Sean Roberson
yesterday












Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
– D. Beec
yesterday





Recall that the chain rule yields that $fracddt f(g(t)) = f'(g(t))*g'(t)$. In your case, we'll call $f(t) = textln (t)$ and let $g(t) = mu (t)$
– D. Beec
yesterday













Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
– Beneschan
yesterday




Oh okay, that makes perfect sense now! I forgot to try the chain rule. Thank you.
– Beneschan
yesterday










3 Answers
3






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It is the chain rule $$(ln(f(x))'=frac1f(x)cdot f'(x)$$ for $$f(x)>0$$






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    Apply chain rule :
    $$(ln( u(t))'=frac ddt ln (u(t))=frac d ln udufrac dudt=frac 1 u(t)frac dudt=frac u'u$$






    share|cite|improve this answer




























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      If we let $$I=intfracf'(x)f(x)dx$$
      then we can use substitution $u=f(x) therefore dx=fracduf'(x)$
      so the integral becomes:
      $$I=intfrac1udu=ln|u|=ln|f(x)|+C therefore fracddxleft[ln(f(x))right]=fracf'(x)f(x)$$






      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








        up vote
        3
        down vote



        accepted










        It is the chain rule $$(ln(f(x))'=frac1f(x)cdot f'(x)$$ for $$f(x)>0$$






        share|cite|improve this answer

























          up vote
          3
          down vote



          accepted










          It is the chain rule $$(ln(f(x))'=frac1f(x)cdot f'(x)$$ for $$f(x)>0$$






          share|cite|improve this answer























            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            It is the chain rule $$(ln(f(x))'=frac1f(x)cdot f'(x)$$ for $$f(x)>0$$






            share|cite|improve this answer













            It is the chain rule $$(ln(f(x))'=frac1f(x)cdot f'(x)$$ for $$f(x)>0$$







            share|cite|improve this answer













            share|cite|improve this answer



            share|cite|improve this answer











            answered yesterday









            Dr. Sonnhard Graubner

            66.6k32659




            66.6k32659




















                up vote
                0
                down vote













                Apply chain rule :
                $$(ln( u(t))'=frac ddt ln (u(t))=frac d ln udufrac dudt=frac 1 u(t)frac dudt=frac u'u$$






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  Apply chain rule :
                  $$(ln( u(t))'=frac ddt ln (u(t))=frac d ln udufrac dudt=frac 1 u(t)frac dudt=frac u'u$$






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Apply chain rule :
                    $$(ln( u(t))'=frac ddt ln (u(t))=frac d ln udufrac dudt=frac 1 u(t)frac dudt=frac u'u$$






                    share|cite|improve this answer













                    Apply chain rule :
                    $$(ln( u(t))'=frac ddt ln (u(t))=frac d ln udufrac dudt=frac 1 u(t)frac dudt=frac u'u$$







                    share|cite|improve this answer













                    share|cite|improve this answer



                    share|cite|improve this answer











                    answered yesterday









                    Isham

                    10.4k3729




                    10.4k3729




















                        up vote
                        0
                        down vote













                        If we let $$I=intfracf'(x)f(x)dx$$
                        then we can use substitution $u=f(x) therefore dx=fracduf'(x)$
                        so the integral becomes:
                        $$I=intfrac1udu=ln|u|=ln|f(x)|+C therefore fracddxleft[ln(f(x))right]=fracf'(x)f(x)$$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote













                          If we let $$I=intfracf'(x)f(x)dx$$
                          then we can use substitution $u=f(x) therefore dx=fracduf'(x)$
                          so the integral becomes:
                          $$I=intfrac1udu=ln|u|=ln|f(x)|+C therefore fracddxleft[ln(f(x))right]=fracf'(x)f(x)$$






                          share|cite|improve this answer























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            If we let $$I=intfracf'(x)f(x)dx$$
                            then we can use substitution $u=f(x) therefore dx=fracduf'(x)$
                            so the integral becomes:
                            $$I=intfrac1udu=ln|u|=ln|f(x)|+C therefore fracddxleft[ln(f(x))right]=fracf'(x)f(x)$$






                            share|cite|improve this answer













                            If we let $$I=intfracf'(x)f(x)dx$$
                            then we can use substitution $u=f(x) therefore dx=fracduf'(x)$
                            so the integral becomes:
                            $$I=intfrac1udu=ln|u|=ln|f(x)|+C therefore fracddxleft[ln(f(x))right]=fracf'(x)f(x)$$







                            share|cite|improve this answer













                            share|cite|improve this answer



                            share|cite|improve this answer











                            answered yesterday









                            Henry Lee

                            48210




                            48210






















                                 

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