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    How would i find the gradient function of this graph

    y=((-x^4+x^2+x+3)/2)^(1/3)

    basically i can't differentiate it and would love someone to help me with it

    thanks
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    differentiate using the chain rule?
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    I dont know the chain rule, im only doing C3 maths at college at the minute.
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    Isn't chain rule at the start of C3?
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    all i have done of c3 at the minute is the coursework which is what i'm needing this for. im showing that the rearrangment method of f(x)=0 to x=g(x) only works when g'(x) is -1<g'(x)<1
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    If you have (F(x))^n , you basically use the standard pattern.

    Ie multiply by the index (n), then subtract 1 from the index.

    Then you differentiate what is in the bracket and mutiply the two results together to give the final result.

    Ie to differentiate (7x)^2, you get 2(7x) and then you differentiate the bracket to get a 7, meaning the final answer is 14(7x) or 98x.
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    y=((-x^4+x^2+x+3)/2)^(1/3) if we remember rules of indices can be written:
    y=((-x^4+x^2+x+3)^-2)^(1/3)
    y=((-x^4+x^2+x+3)^(-2)x(1/3)
    y=((-x^4+x^2+x+3)^-2/3

    now in form (f(x)^n)
    chain rule dy/dx= (n)(f'(x))(f(x))^n-1)
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    (Original post by stevencarrwork)
    If you have (F(x))^n , you basically use the standard pattern.

    Ie multiply by the index (n), then subtract 1 from the index.

    Then you differentiate what is in the bracket and mutiply the two results together to give the final result.

    Ie to differentiate (7x)^2, you get 2(7x) and then you differentiate the bracket to get a 7, meaning the final answer is 14(7x) or 98x.
    Okay, tried my hardest to understand/follow this but i dont quite get it.

    I know i wont learn anything from this, but please could someone just give me the answer as i need to finish writing the courswork tonight and this the very last section.
    thanks
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    (Original post by 1k0ed1)
    How would i find the gradient function of this graph

    y=((-x^4+x^2+x+3)/2)^(1/3)

    basically i can't differentiate it and would love someone to help me with it

    thanks
    \lett(\dfrac{-x^4+x^2+x+3}{2}\right)^\frac{1}{  3}

    I can't get the right bracket. Help
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    (Original post by davidcy147)
    y=((-x^4+x^2+x+3)/2)^(1/3) if we remember rules of indices can be written:
    y=((-x^4+x^2+x+3)^-2)^(1/3)
    y=((-x^4+x^2+x+3)^(-2)x(1/3)
    y=((-x^4+x^2+x+3)^-2/3

    now in form (f(x)^n)
    chain rule dy/dx= (n-1)(f'(x))(f(x))^n)
    Does this not help because if you can differentiate just use this chain rule here. PM if you need help
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    chain Rule [email protected]
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    (Original post by 1k0ed1)
    How would i find the gradient function of this graph

    y=((-x^4+x^2+x+3)/2)^(1/3)

    basically i can't differentiate it and would love someone to help me with it

    thanks
    I think this might be helpful.
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    (Original post by maxfire)
    chain Rule [email protected]
    havnt learnt it yet :p:
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    diff the bracket

    bring the power to the front

    subtract 1 from the power

    then you have (power)(diff of bracket)(bracket)^(power - 1)

    (correct me if I'm wrong)
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    Cubing it and differentiating implicitly would also be a cool way to do this!

    EDIT: More generally: \displaystyle \frac{d}{dx}\left f(g(x)) \right = f'(g(x))g'(x)
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    (Original post by My Alt)
    Cubing it and differentiating implicitly would also be a cool way to do this!
    Seriously! ...Why would you? STEP may interest you! :p:
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    (Original post by 1k0ed1)
    havnt learnt it yet :p:

    erg.
    ok
    if you have something in a bracket (u) to the power of something (n)

    u^n

    differentiates to:



    n(u')(u)^(n-1)

    if you get that.

    so you bring the power down, and minus 1 from it like usual, and then you time it by the derivative of the funtion as well (u')


    HOPE THIS HELPED
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    (Original post by davidcy147)
    Seriously! ...Why would you? STEP may interest you! :p:
    It's not really any slower, and it means you dont _need_ the chain rule :p: What do you mean STEP may interest you?!?
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    Haha Step is all about thinking outside the box, which is something you seem to do! Kudos to you though, wish I had that creative thought pattern!
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    (Original post by My Alt)
    Cubing it and differentiating implicitly would also be a cool way to do this!

    EDIT: More generally: \displaystyle \frac{d}{dx}\left f(g(x)) \right = f'(g(x))g'(x)
    i wish i understood this
 
 
 
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