Why can't I use the formula for $f(x)=x^x$ to derive $(frac{x}{x+1})^x$?












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Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






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    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10
















0












$begingroup$


Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10














0












0








0





$begingroup$


Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives







share|cite|improve this question











$endgroup$




Why can't I use the formula $y=x^x$, $dot {y}=x^x(ln x+1)$ to derive $(frac{x}{x+1})^x$?
I've tried and it doesn't get me the correct result. Have I done any mistake?
$$Bigl[Bigl(frac{x}{x+1}Bigr)^xBigr]'=(frac{x}{x+1})^xBigl[lnBigl(frac{x}{x+1}Bigr)+1Bigr]Bigl[frac{1+x-x}{(x+1)^2}Bigr]=frac{bigl(frac{x}{x+1}bigr)^xBigl[lnbigl(frac{x}{x+1}bigr)+1Bigr]}{(1+x)^2}$$
If not, please explain why I can't use the formula.






derivatives




derivatives






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edited Jan 10 at 19:14









Bernard

122k740116




122k740116










asked Jan 10 at 19:10









Lauren SinLauren Sin

1055




1055








  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10














  • 2




    $begingroup$
    Simple: you need the quotient and chain rules.
    $endgroup$
    – Sean Roberson
    Jan 10 at 19:10








2




2




$begingroup$
Simple: you need the quotient and chain rules.
$endgroup$
– Sean Roberson
Jan 10 at 19:10




$begingroup$
Simple: you need the quotient and chain rules.
$endgroup$
– Sean Roberson
Jan 10 at 19:10










5 Answers
5






active

oldest

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2












$begingroup$

Hint:



Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
$$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
+x,frac{u'(x)}{u(x)}.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
    $endgroup$
    – Bernard
    Jan 10 at 20:40



















2












$begingroup$

Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
      left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
      right) +1 right) }
      $$
      Taking the logarithm on both sides we get
      $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
      $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






      share|cite|improve this answer











      $endgroup$





















        1












        $begingroup$

        What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






        share|cite|improve this answer









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          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40
















          2












          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40














          2












          2








          2





          $begingroup$

          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$






          share|cite|improve this answer











          $endgroup$



          Hint:



          Your error comes from the fact that in your formula, the variable and the exponent have to be identical. I suggest using the logarithmic derivative: if $f(x)=bigl(u(x)bigr)^x$
          $$frac{f'(x)}{f(x)}=lnbigl(u(x)bigr)
          +x,frac{u'(x)}{u(x)}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 20:39

























          answered Jan 10 at 19:32









          BernardBernard

          122k740116




          122k740116












          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40


















          • $begingroup$
            Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
            $endgroup$
            – Bernard
            Jan 10 at 20:40
















          $begingroup$
          Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
          $endgroup$
          – Bernard
          Jan 10 at 20:40




          $begingroup$
          Sorry for the typo – I forgot a factor $x$. The formula is fixed now.
          $endgroup$
          – Bernard
          Jan 10 at 20:40











          2












          $begingroup$

          Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






          share|cite|improve this answer









          $endgroup$


















            2












            $begingroup$

            Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






            share|cite|improve this answer









            $endgroup$
















              2












              2








              2





              $begingroup$

              Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.






              share|cite|improve this answer









              $endgroup$



              Define $z:=frac{x}{x+1}$. Your claim is that $frac{d}{dx}(z^x)=frac{d}{dz}(z^z)cdotfrac{dz}{dx}$, which is clearly wrong because the right-hand side is actually $frac{d}{dx}(z^z)$. I'll leave you to solve the problem properly, the very same way you'd differentiate $x^x$ in the first place: by logarithmic differentiation.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 10 at 19:16









              J.G.J.G.

              28.7k22845




              28.7k22845























                  2












                  $begingroup$

                  To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                  share|cite|improve this answer









                  $endgroup$


















                    2












                    $begingroup$

                    To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                    share|cite|improve this answer









                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.






                      share|cite|improve this answer









                      $endgroup$



                      To apply the chain rule directly, you need something of the form $f(g(x))$; in your case that would be $g(x)^{g(x)}$, but that's not what you have: you have $g(x)^x$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 10 at 19:22









                      Martin ArgeramiMartin Argerami

                      128k1183183




                      128k1183183























                          1












                          $begingroup$

                          Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                          left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                          right) +1 right) }
                          $$
                          Taking the logarithm on both sides we get
                          $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                          $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                            left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                            right) +1 right) }
                            $$
                            Taking the logarithm on both sides we get
                            $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                            $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                              left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                              right) +1 right) }
                              $$
                              Taking the logarithm on both sides we get
                              $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                              $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$






                              share|cite|improve this answer











                              $endgroup$



                              Hint: I have got this here :$${frac {1}{x+1} left( {frac {x}{x+1}} right) ^{x} left( ln
                              left( {frac {x}{x+1}} right) x+ln left( {frac {x}{x+1}}
                              right) +1 right) }
                              $$
                              Taking the logarithm on both sides we get
                              $$ln(y)=x(ln(x)-ln(x+1))$$ and differentiating with respect to $x$
                              $$frac{1}{y}y'=ln(x)-ln(x+1)+xleft(frac{1}{x}-frac{1}{x+1}right)$$







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Jan 10 at 19:34

























                              answered Jan 10 at 19:14









                              Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                              76.8k42866




                              76.8k42866























                                  1












                                  $begingroup$

                                  What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$






                                      share|cite|improve this answer









                                      $endgroup$



                                      What you have used is that $$left [g(x)^xright]'=g'(x)cdot g(x)^xcdot ln g(x)+1$$which doesn't generally hold. To solve this problem just simply write$$ln f(x)=xln x-xln(x+1)$$whose first-order differentiation leads to $${f'(x)over f(x)}=1+ln x-ln (x+1)-{xover x+1}$$from which we obtain$$f'(x)=left({xover x+1}right)^xcdot left({1over x+1}+ln{xover x+1}right)$$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 10 at 20:19









                                      Mostafa AyazMostafa Ayaz

                                      15.7k3939




                                      15.7k3939






























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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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