Differentiate the exponential function $f(x)= frac{x^2e^x}{x^2+e^x}$












0












$begingroup$


$$f(x)= frac{x^2e^x}{x^2+e^x}$$



Using product rule and quotient rule I computed



$$f'(x)=frac{(x^2+e^x)e^x(x^2 + 2 x ) - x^2e^x(2x+e^x)}{(x^2+e^x)^2}$$



Is my computation correct so far?






calculus exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The 1st part of the numerator is computed wrongly I think.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:57










  • $begingroup$
    Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 0:57










  • $begingroup$
    @zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:02












  • $begingroup$
    Yes. That's right :)
    $endgroup$
    – mm-crj
    Dec 11 '18 at 1:05
















0












$begingroup$


$$f(x)= frac{x^2e^x}{x^2+e^x}$$



Using product rule and quotient rule I computed



$$f'(x)=frac{(x^2+e^x)e^x(x^2 + 2 x ) - x^2e^x(2x+e^x)}{(x^2+e^x)^2}$$



Is my computation correct so far?






calculus exponential-function







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The 1st part of the numerator is computed wrongly I think.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:57










  • $begingroup$
    Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 0:57










  • $begingroup$
    @zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:02












  • $begingroup$
    Yes. That's right :)
    $endgroup$
    – mm-crj
    Dec 11 '18 at 1:05














0












0








0





$begingroup$


$$f(x)= frac{x^2e^x}{x^2+e^x}$$



Using product rule and quotient rule I computed



$$f'(x)=frac{(x^2+e^x)e^x(x^2 + 2 x ) - x^2e^x(2x+e^x)}{(x^2+e^x)^2}$$



Is my computation correct so far?






calculus exponential-function







share|cite|improve this question











$endgroup$




$$f(x)= frac{x^2e^x}{x^2+e^x}$$



Using product rule and quotient rule I computed



$$f'(x)=frac{(x^2+e^x)e^x(x^2 + 2 x ) - x^2e^x(2x+e^x)}{(x^2+e^x)^2}$$



Is my computation correct so far?






calculus exponential-function




calculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 3:02









Key Flex

8,28261233




8,28261233










asked Dec 11 '18 at 0:41









Eric BrownEric Brown

757




757








  • 1




    $begingroup$
    The 1st part of the numerator is computed wrongly I think.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:57










  • $begingroup$
    Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 0:57










  • $begingroup$
    @zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:02












  • $begingroup$
    Yes. That's right :)
    $endgroup$
    – mm-crj
    Dec 11 '18 at 1:05














  • 1




    $begingroup$
    The 1st part of the numerator is computed wrongly I think.
    $endgroup$
    – mm-crj
    Dec 11 '18 at 0:57










  • $begingroup$
    Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 0:57










  • $begingroup$
    @zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:02












  • $begingroup$
    Yes. That's right :)
    $endgroup$
    – mm-crj
    Dec 11 '18 at 1:05








1




1




$begingroup$
The 1st part of the numerator is computed wrongly I think.
$endgroup$
– mm-crj
Dec 11 '18 at 0:57




$begingroup$
The 1st part of the numerator is computed wrongly I think.
$endgroup$
– mm-crj
Dec 11 '18 at 0:57












$begingroup$
Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
$endgroup$
– AryanSonwatikar
Dec 11 '18 at 0:57




$begingroup$
Yes, except the fact that you will have to put brackets around $x^2 + e^x$ in the numerator.
$endgroup$
– AryanSonwatikar
Dec 11 '18 at 0:57












$begingroup$
@zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
$endgroup$
– AryanSonwatikar
Dec 11 '18 at 1:02






$begingroup$
@zenith, right. It should be $(x^2+e^x)e^x(x^2+2x)$
$endgroup$
– AryanSonwatikar
Dec 11 '18 at 1:02














$begingroup$
Yes. That's right :)
$endgroup$
– mm-crj
Dec 11 '18 at 1:05




$begingroup$
Yes. That's right :)
$endgroup$
– mm-crj
Dec 11 '18 at 1:05










2 Answers
2






active

oldest

votes


















2












$begingroup$

Given $dfrac{x^2e^x}{x^2+e^x}$



Apply the quotient rule: $left(dfrac uvright)=dfrac{u^{primecdot}cdot v-v^{prime}cdot u}{v^2}$
$$dfrac{d}{dx}left(dfrac{x^2e^x}{x^2+e^x}right)=dfrac{dfrac{d}{dx}(x^2e^x)(x^2+e^x)-dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$



$$=dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$



Edit:



$dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule



By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 1:13












  • $begingroup$
    I don't understand what you have done. (P.S. I am a calculus beginner.)
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:13










  • $begingroup$
    @EricBrown they are the same thing.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:14










  • $begingroup$
    @AryanSonwatikar Is it clear now?
    $endgroup$
    – Key Flex
    Dec 11 '18 at 1:26










  • $begingroup$
    @KeyFlex Yes, it is.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:50



















1












$begingroup$

If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have
$$ln f(x) = 2 ln x + x - ln (x^2 + e^x).$$
Differentiating with respect to $x$ gives
$$frac{f'(x)}{f(x)} = frac{2}{x} + 1 - frac{2x + e^x}{x^2 + e^x},$$
which, after some algebra and simplification, reduces to
$$f'(x) = frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    2












    $begingroup$

    Given $dfrac{x^2e^x}{x^2+e^x}$



    Apply the quotient rule: $left(dfrac uvright)=dfrac{u^{primecdot}cdot v-v^{prime}cdot u}{v^2}$
    $$dfrac{d}{dx}left(dfrac{x^2e^x}{x^2+e^x}right)=dfrac{dfrac{d}{dx}(x^2e^x)(x^2+e^x)-dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$



    $$=dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$



    Edit:



    $dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule



    By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
      $endgroup$
      – Eric Brown
      Dec 11 '18 at 1:13












    • $begingroup$
      I don't understand what you have done. (P.S. I am a calculus beginner.)
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:13










    • $begingroup$
      @EricBrown they are the same thing.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:14










    • $begingroup$
      @AryanSonwatikar Is it clear now?
      $endgroup$
      – Key Flex
      Dec 11 '18 at 1:26










    • $begingroup$
      @KeyFlex Yes, it is.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:50
















    2












    $begingroup$

    Given $dfrac{x^2e^x}{x^2+e^x}$



    Apply the quotient rule: $left(dfrac uvright)=dfrac{u^{primecdot}cdot v-v^{prime}cdot u}{v^2}$
    $$dfrac{d}{dx}left(dfrac{x^2e^x}{x^2+e^x}right)=dfrac{dfrac{d}{dx}(x^2e^x)(x^2+e^x)-dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$



    $$=dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$



    Edit:



    $dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule



    By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
      $endgroup$
      – Eric Brown
      Dec 11 '18 at 1:13












    • $begingroup$
      I don't understand what you have done. (P.S. I am a calculus beginner.)
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:13










    • $begingroup$
      @EricBrown they are the same thing.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:14










    • $begingroup$
      @AryanSonwatikar Is it clear now?
      $endgroup$
      – Key Flex
      Dec 11 '18 at 1:26










    • $begingroup$
      @KeyFlex Yes, it is.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:50














    2












    2








    2





    $begingroup$

    Given $dfrac{x^2e^x}{x^2+e^x}$



    Apply the quotient rule: $left(dfrac uvright)=dfrac{u^{primecdot}cdot v-v^{prime}cdot u}{v^2}$
    $$dfrac{d}{dx}left(dfrac{x^2e^x}{x^2+e^x}right)=dfrac{dfrac{d}{dx}(x^2e^x)(x^2+e^x)-dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$



    $$=dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$



    Edit:



    $dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule



    By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$






    share|cite|improve this answer











    $endgroup$



    Given $dfrac{x^2e^x}{x^2+e^x}$



    Apply the quotient rule: $left(dfrac uvright)=dfrac{u^{primecdot}cdot v-v^{prime}cdot u}{v^2}$
    $$dfrac{d}{dx}left(dfrac{x^2e^x}{x^2+e^x}right)=dfrac{dfrac{d}{dx}(x^2e^x)(x^2+e^x)-dfrac{d}{dx}(x^2+e^x)x^2e^x}{(x^2+e^x)^2}=dfrac{(2xe^x+e^xx^2)(x^2+e^x)-(2x+e^x)(x^2e^x)}{(x^2+e^x)^2}$$



    $$=dfrac{e^x+x^4+2e^{2x}x}{(x^2+e^x)^2}$$



    Edit:



    $dfrac{d}{dx}(x^2e^x)=2xe^x+e^xx^2$ by using product rule



    By expanding $(2xe^x+e^x x^2)(x^2+e^x)-(2x+e^x)(x^2e^x)$ we get $e^xx^4+2e^{2x}x$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 1:26

























    answered Dec 11 '18 at 1:05









    Key FlexKey Flex

    8,28261233




    8,28261233












    • $begingroup$
      On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
      $endgroup$
      – Eric Brown
      Dec 11 '18 at 1:13












    • $begingroup$
      I don't understand what you have done. (P.S. I am a calculus beginner.)
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:13










    • $begingroup$
      @EricBrown they are the same thing.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:14










    • $begingroup$
      @AryanSonwatikar Is it clear now?
      $endgroup$
      – Key Flex
      Dec 11 '18 at 1:26










    • $begingroup$
      @KeyFlex Yes, it is.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:50


















    • $begingroup$
      On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
      $endgroup$
      – Eric Brown
      Dec 11 '18 at 1:13












    • $begingroup$
      I don't understand what you have done. (P.S. I am a calculus beginner.)
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:13










    • $begingroup$
      @EricBrown they are the same thing.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:14










    • $begingroup$
      @AryanSonwatikar Is it clear now?
      $endgroup$
      – Key Flex
      Dec 11 '18 at 1:26










    • $begingroup$
      @KeyFlex Yes, it is.
      $endgroup$
      – AryanSonwatikar
      Dec 11 '18 at 1:50
















    $begingroup$
    On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 1:13






    $begingroup$
    On the 2nd line that $2xe^x+e^xx^2$, shouldn't it be $2xe^x+x^2e^x$?
    $endgroup$
    – Eric Brown
    Dec 11 '18 at 1:13














    $begingroup$
    I don't understand what you have done. (P.S. I am a calculus beginner.)
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:13




    $begingroup$
    I don't understand what you have done. (P.S. I am a calculus beginner.)
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:13












    $begingroup$
    @EricBrown they are the same thing.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:14




    $begingroup$
    @EricBrown they are the same thing.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:14












    $begingroup$
    @AryanSonwatikar Is it clear now?
    $endgroup$
    – Key Flex
    Dec 11 '18 at 1:26




    $begingroup$
    @AryanSonwatikar Is it clear now?
    $endgroup$
    – Key Flex
    Dec 11 '18 at 1:26












    $begingroup$
    @KeyFlex Yes, it is.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:50




    $begingroup$
    @KeyFlex Yes, it is.
    $endgroup$
    – AryanSonwatikar
    Dec 11 '18 at 1:50











    1












    $begingroup$

    If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have
    $$ln f(x) = 2 ln x + x - ln (x^2 + e^x).$$
    Differentiating with respect to $x$ gives
    $$frac{f'(x)}{f(x)} = frac{2}{x} + 1 - frac{2x + e^x}{x^2 + e^x},$$
    which, after some algebra and simplification, reduces to
    $$f'(x) = frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have
      $$ln f(x) = 2 ln x + x - ln (x^2 + e^x).$$
      Differentiating with respect to $x$ gives
      $$frac{f'(x)}{f(x)} = frac{2}{x} + 1 - frac{2x + e^x}{x^2 + e^x},$$
      which, after some algebra and simplification, reduces to
      $$f'(x) = frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have
        $$ln f(x) = 2 ln x + x - ln (x^2 + e^x).$$
        Differentiating with respect to $x$ gives
        $$frac{f'(x)}{f(x)} = frac{2}{x} + 1 - frac{2x + e^x}{x^2 + e^x},$$
        which, after some algebra and simplification, reduces to
        $$f'(x) = frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$






        share|cite|improve this answer









        $endgroup$



        If you would like to avoid the product and quotient rules altogether you could take the natural logarithm of both sides before differentiating. In doing so we have
        $$ln f(x) = 2 ln x + x - ln (x^2 + e^x).$$
        Differentiating with respect to $x$ gives
        $$frac{f'(x)}{f(x)} = frac{2}{x} + 1 - frac{2x + e^x}{x^2 + e^x},$$
        which, after some algebra and simplification, reduces to
        $$f'(x) = frac{x e^x (2e^x + x^3)}{(x^2 + e^x)^2}.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 2:02









        omegadotomegadot

        6,0222828




        6,0222828






























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            Plaza Victoria

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            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

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            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
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            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

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            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
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