The derivative of $ 2^{frac{x}{ln x}} $












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I am supposed to find the derivative of $ 2^{frac{x}{ln x}} $. My answer is $$ 2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}cdot frac{1}{x} .$$ Is it correct? Thanks.






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    I am supposed to find the derivative of $ 2^{frac{x}{ln x}} $. My answer is $$ 2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}cdot frac{1}{x} .$$ Is it correct? Thanks.






    derivatives







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      I am supposed to find the derivative of $ 2^{frac{x}{ln x}} $. My answer is $$ 2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}cdot frac{1}{x} .$$ Is it correct? Thanks.






      derivatives







      share|cite|improve this question















      I am supposed to find the derivative of $ 2^{frac{x}{ln x}} $. My answer is $$ 2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}cdot frac{1}{x} .$$ Is it correct? Thanks.






      derivatives




      derivatives






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      edited Nov 26 '18 at 19:49









      Robert Z

      93.3k1061132




      93.3k1061132










      asked Nov 26 '18 at 19:05









      Johny547

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          3 Answers
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          You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is
          $$D(2^{frac{x}{ln x}})=2^{frac{x}{ln x}} cdot ln 2 cdot Dleft(frac{x}{ln x}right)=2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}.$$






          share|cite|improve this answer























          • Isn't your answer the same as mine?
            – Johny547
            Nov 26 '18 at 19:20










          • Your answer had an extra $frac{1}{x}$ at the end.
            – KM101
            Nov 26 '18 at 19:22










          • No, you have en extra factor $1/x$ ate the end.
            – Robert Z
            Nov 26 '18 at 19:22










          • Thanks, I see it now.
            – Johny547
            Nov 26 '18 at 19:23



















          1














          Let $y=2^frac{x}{ln{x}}$. Then



          $$ln{y}=ln{2}^{frac{x}{ln{x}}}=frac{x}{ln{x}}cdotln2$$



          Now $$frac{1}{y}cdotfrac{dy}{dx}=ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



          Now multiply by $y$ and get



          $$frac{dy}{dx}=yln{2}left[frac{ln{x}-1}{(ln x)^2}right]=2^frac{x}{ln{x}}ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



          You can do all sorts of algebra to reduce this but it is the derivative.






          share|cite|improve this answer





























            0














            Mathematica gives:



            $$frac{log (2) 2^{frac{x}{log (x)}} (log (x)-1)}{log ^2(x)}$$






            share|cite|improve this answer























            • But that is after simplification, right? Mine is without.
              – Johny547
              Nov 26 '18 at 19:16












            • Did you press the right buttons? Your derivative seems to be wrong!
              – Robert Z
              Nov 26 '18 at 19:30












            • Oopsss... fixed. Thanks.
              – David G. Stork
              Nov 26 '18 at 19:32











            Your Answer





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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

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            active

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            1














            You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is
            $$D(2^{frac{x}{ln x}})=2^{frac{x}{ln x}} cdot ln 2 cdot Dleft(frac{x}{ln x}right)=2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}.$$






            share|cite|improve this answer























            • Isn't your answer the same as mine?
              – Johny547
              Nov 26 '18 at 19:20










            • Your answer had an extra $frac{1}{x}$ at the end.
              – KM101
              Nov 26 '18 at 19:22










            • No, you have en extra factor $1/x$ ate the end.
              – Robert Z
              Nov 26 '18 at 19:22










            • Thanks, I see it now.
              – Johny547
              Nov 26 '18 at 19:23
















            1














            You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is
            $$D(2^{frac{x}{ln x}})=2^{frac{x}{ln x}} cdot ln 2 cdot Dleft(frac{x}{ln x}right)=2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}.$$






            share|cite|improve this answer























            • Isn't your answer the same as mine?
              – Johny547
              Nov 26 '18 at 19:20










            • Your answer had an extra $frac{1}{x}$ at the end.
              – KM101
              Nov 26 '18 at 19:22










            • No, you have en extra factor $1/x$ ate the end.
              – Robert Z
              Nov 26 '18 at 19:22










            • Thanks, I see it now.
              – Johny547
              Nov 26 '18 at 19:23














            1












            1








            1






            You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is
            $$D(2^{frac{x}{ln x}})=2^{frac{x}{ln x}} cdot ln 2 cdot Dleft(frac{x}{ln x}right)=2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}.$$






            share|cite|improve this answer














            You are almost correct. You have just an extra factor $1/x$ at the end. The correct derivative is
            $$D(2^{frac{x}{ln x}})=2^{frac{x}{ln x}} cdot ln 2 cdot Dleft(frac{x}{ln x}right)=2^{frac{x}{ln x}} cdot ln 2 cdot frac{ln x-xcdot frac{1}{x}}{ln^{2}x}.$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 26 '18 at 19:56

























            answered Nov 26 '18 at 19:19









            Robert Z

            93.3k1061132




            93.3k1061132












            • Isn't your answer the same as mine?
              – Johny547
              Nov 26 '18 at 19:20










            • Your answer had an extra $frac{1}{x}$ at the end.
              – KM101
              Nov 26 '18 at 19:22










            • No, you have en extra factor $1/x$ ate the end.
              – Robert Z
              Nov 26 '18 at 19:22










            • Thanks, I see it now.
              – Johny547
              Nov 26 '18 at 19:23


















            • Isn't your answer the same as mine?
              – Johny547
              Nov 26 '18 at 19:20










            • Your answer had an extra $frac{1}{x}$ at the end.
              – KM101
              Nov 26 '18 at 19:22










            • No, you have en extra factor $1/x$ ate the end.
              – Robert Z
              Nov 26 '18 at 19:22










            • Thanks, I see it now.
              – Johny547
              Nov 26 '18 at 19:23
















            Isn't your answer the same as mine?
            – Johny547
            Nov 26 '18 at 19:20




            Isn't your answer the same as mine?
            – Johny547
            Nov 26 '18 at 19:20












            Your answer had an extra $frac{1}{x}$ at the end.
            – KM101
            Nov 26 '18 at 19:22




            Your answer had an extra $frac{1}{x}$ at the end.
            – KM101
            Nov 26 '18 at 19:22












            No, you have en extra factor $1/x$ ate the end.
            – Robert Z
            Nov 26 '18 at 19:22




            No, you have en extra factor $1/x$ ate the end.
            – Robert Z
            Nov 26 '18 at 19:22












            Thanks, I see it now.
            – Johny547
            Nov 26 '18 at 19:23




            Thanks, I see it now.
            – Johny547
            Nov 26 '18 at 19:23











            1














            Let $y=2^frac{x}{ln{x}}$. Then



            $$ln{y}=ln{2}^{frac{x}{ln{x}}}=frac{x}{ln{x}}cdotln2$$



            Now $$frac{1}{y}cdotfrac{dy}{dx}=ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



            Now multiply by $y$ and get



            $$frac{dy}{dx}=yln{2}left[frac{ln{x}-1}{(ln x)^2}right]=2^frac{x}{ln{x}}ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



            You can do all sorts of algebra to reduce this but it is the derivative.






            share|cite|improve this answer


























              1














              Let $y=2^frac{x}{ln{x}}$. Then



              $$ln{y}=ln{2}^{frac{x}{ln{x}}}=frac{x}{ln{x}}cdotln2$$



              Now $$frac{1}{y}cdotfrac{dy}{dx}=ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



              Now multiply by $y$ and get



              $$frac{dy}{dx}=yln{2}left[frac{ln{x}-1}{(ln x)^2}right]=2^frac{x}{ln{x}}ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



              You can do all sorts of algebra to reduce this but it is the derivative.






              share|cite|improve this answer
























                1












                1








                1






                Let $y=2^frac{x}{ln{x}}$. Then



                $$ln{y}=ln{2}^{frac{x}{ln{x}}}=frac{x}{ln{x}}cdotln2$$



                Now $$frac{1}{y}cdotfrac{dy}{dx}=ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



                Now multiply by $y$ and get



                $$frac{dy}{dx}=yln{2}left[frac{ln{x}-1}{(ln x)^2}right]=2^frac{x}{ln{x}}ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



                You can do all sorts of algebra to reduce this but it is the derivative.






                share|cite|improve this answer












                Let $y=2^frac{x}{ln{x}}$. Then



                $$ln{y}=ln{2}^{frac{x}{ln{x}}}=frac{x}{ln{x}}cdotln2$$



                Now $$frac{1}{y}cdotfrac{dy}{dx}=ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



                Now multiply by $y$ and get



                $$frac{dy}{dx}=yln{2}left[frac{ln{x}-1}{(ln x)^2}right]=2^frac{x}{ln{x}}ln{2}left[frac{ln{x}-1}{(ln x)^2}right]$$



                You can do all sorts of algebra to reduce this but it is the derivative.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 '18 at 19:17









                Eleven-Eleven

                5,39072659




                5,39072659























                    0














                    Mathematica gives:



                    $$frac{log (2) 2^{frac{x}{log (x)}} (log (x)-1)}{log ^2(x)}$$






                    share|cite|improve this answer























                    • But that is after simplification, right? Mine is without.
                      – Johny547
                      Nov 26 '18 at 19:16












                    • Did you press the right buttons? Your derivative seems to be wrong!
                      – Robert Z
                      Nov 26 '18 at 19:30












                    • Oopsss... fixed. Thanks.
                      – David G. Stork
                      Nov 26 '18 at 19:32
















                    0














                    Mathematica gives:



                    $$frac{log (2) 2^{frac{x}{log (x)}} (log (x)-1)}{log ^2(x)}$$






                    share|cite|improve this answer























                    • But that is after simplification, right? Mine is without.
                      – Johny547
                      Nov 26 '18 at 19:16












                    • Did you press the right buttons? Your derivative seems to be wrong!
                      – Robert Z
                      Nov 26 '18 at 19:30












                    • Oopsss... fixed. Thanks.
                      – David G. Stork
                      Nov 26 '18 at 19:32














                    0












                    0








                    0






                    Mathematica gives:



                    $$frac{log (2) 2^{frac{x}{log (x)}} (log (x)-1)}{log ^2(x)}$$






                    share|cite|improve this answer














                    Mathematica gives:



                    $$frac{log (2) 2^{frac{x}{log (x)}} (log (x)-1)}{log ^2(x)}$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 26 '18 at 19:32

























                    answered Nov 26 '18 at 19:15









                    David G. Stork

                    9,82021232




                    9,82021232












                    • But that is after simplification, right? Mine is without.
                      – Johny547
                      Nov 26 '18 at 19:16












                    • Did you press the right buttons? Your derivative seems to be wrong!
                      – Robert Z
                      Nov 26 '18 at 19:30












                    • Oopsss... fixed. Thanks.
                      – David G. Stork
                      Nov 26 '18 at 19:32


















                    • But that is after simplification, right? Mine is without.
                      – Johny547
                      Nov 26 '18 at 19:16












                    • Did you press the right buttons? Your derivative seems to be wrong!
                      – Robert Z
                      Nov 26 '18 at 19:30












                    • Oopsss... fixed. Thanks.
                      – David G. Stork
                      Nov 26 '18 at 19:32
















                    But that is after simplification, right? Mine is without.
                    – Johny547
                    Nov 26 '18 at 19:16






                    But that is after simplification, right? Mine is without.
                    – Johny547
                    Nov 26 '18 at 19:16














                    Did you press the right buttons? Your derivative seems to be wrong!
                    – Robert Z
                    Nov 26 '18 at 19:30






                    Did you press the right buttons? Your derivative seems to be wrong!
                    – Robert Z
                    Nov 26 '18 at 19:30














                    Oopsss... fixed. Thanks.
                    – David G. Stork
                    Nov 26 '18 at 19:32




                    Oopsss... fixed. Thanks.
                    – David G. Stork
                    Nov 26 '18 at 19:32


















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