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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            Puebla de Zaragoza

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            Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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            Musa

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            Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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