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    I've been trying to figure out where I've gone wrong for ages!! Can someone pls help, thank u




    ---CA Edited post so image was correct way up---
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    Name:  IMG_20170209_131326.jpg
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    Ugh idk how to rotate it.. I guess this ones better
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    e^2x+2x^2e^2x
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    (Original post by DirtyDad)
    e^2x+2x^2e^2x
    what about the lne^x^2 that goes at the end!!
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    When you differentiate e^x^2, there's no need to multiply by another x^2.
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    (Original post by pondsteps)
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    Ugh idk how to rotate it.. I guess this ones better
    I did it for you - check your post

    I recommend using imgur to upload images then post the link (including the file type e.g. ,jpg) inside [img] tags.
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    (Original post by DirtyDad)
    e^2x+2x^2e^2x
    crappp i just realised thats one lol thank u
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    (Original post by pondsteps)
    what about the lne^x^2 that goes at the end!!
    what are you on about LN, there is no need for ln where your intergrating xe^2x
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    (Original post by notnek)
    I did it for you - check your post

    I recommend using imgur to upload images then post the link (including the file type e.g. ,jpg) inside [img] tags.
    okay thank u!
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    (Original post by DirtyDad)
    what are you on about LN, there is no need for ln where your intergrating xe^2x
    this was dy/dx so not really intergrating ?
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    (Original post by the bear)
    this was dy/dx so not really intergrating ?
    Sorry Hard, mistakes happen you know? like when hyperthreading my blackjack game, things just tend to fall apart
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    (Original post by pondsteps)
    I've been trying to figure out where I've gone wrong for ages!! Can someone pls help, thank u


    ---CA Edited post so image was correct way up---

    Using the product rule y=xe^x2
    Let u= x and v= e^x2. Therefore, u'= 1 and v'= 2xe^x2

    So using the product rule dy/dx= u.v' + v.u'
    This mean dy/dx= (x) . (2xe^x2) + (e^x2) . (1)
    This is equal to 2x2e^x2 + e^x2
    Factorising this produces e^x2(2x2 + 1), hope this helps simplify it a little.
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    Maz A thanks so much ))
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    (Original post by pondsteps)
    Maz A thanks so much ))
    Any chance of a rep..? Jks you're very welcome glad it helped .
 
 
 
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