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Math Help - Differentiate

  1. #1
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    Differentiate

    hello everyone,

    I have to differentiate this using the compsite rule only:
    f(x)=e^x(4-x)/6

    I am using k'(x)=g'(f(x)) f'(x)

    where k(x)=g(f(x)) and

    u = f(x) = x(4-x)/6 ,which I multiplied out to give 2/3x - 1/6x^2

    so, f'(x) =2/3 - 2/3x

    g(u)=e^u

    so, k'(x) = (e^u) (2/3 - 2/3x)

    stuck from there, also not sure if all of the above is right. Help please.

    sweeties
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  2. #2
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    Quote Originally Posted by Sweeties View Post
    hello everyone,

    I have to differentiate this using the compsite rule only:
    f(x)=e^x(4-x)/6

    I am using k'(x)=g'(f(x)) f'(x)

    where k(x)=g(f(x)) and

    u = f(x) = x(4-x)/6 ,which I multiplied out to give 2/3x - 1/6x^2

    so, f'(x) =2/3 - 1/3x << SiMoon says : it's 2/6 x=1/3, not 2/3

    g(u)=e^u

    so, k'(x) = (e^u) (2/3 - 2/3x)

    stuck from there, also not sure if all of the above is right. Help please.

    sweeties
    k'(x) = (e^u) (2/3 - 1/3x)

    And u=x(4-x)/6, remember
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  3. #3
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    Quote Originally Posted by Moo View Post
    k'(x) = (e^u) (2/3 - 1/3x)

    And u=x(4-x)/6, remember
    Yeh, i know, but i couldn't multiply it out
    not sure how to do it.
    i had a go and got something horrid like:

    2/3 e ^x(4-x)/6 - 1/3 ex ^ x(4-x)/6
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  4. #4
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    Quote Originally Posted by Sweeties View Post
    Yeh, i know, but i couldn't multiply it out
    not sure how to do it.
    i had a go and got something horrid like:

    2/3 e ^x(4-x)/6 - 1/3 ex ^ x(4-x)/6
    The red thing means that it's e*x^(...), which is false. It's x*e^(...)

    Take care about the order.
    Plus, use parenthesis... You should write e^(x(4-x)/6), so that it's correct.

    Keep it in the form (\frac 23-\frac 13 x)e^{x(4-x)/6}

    It's not so horrible this way
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  5. #5
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    oooh, you're a darling moo!
    i think i have just been studying too long today, need some sunshine!
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  6. #6
    Moo
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    Quote Originally Posted by Sweeties View Post
    oooh, you're a darling moo!
    i think i have just been studying too long today, need some sunshine!
    Never study too long at once, get some fresh air :



    enjoy
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  7. #7
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    i forgot the second part:


    use the product rule to differentiate the same original function and use previous answer to show that g(x) =xe^(x(4-x)/6) has the derivative
    g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6)

    i really dont know how to do this, i tried but got nowhere.
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  8. #8
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    Quote Originally Posted by Sweeties View Post
    i forgot the second part:


    use the product rule to differentiate the same original function and use previous answer to show that g(x) =xe^(x(4-x)/6) has the derivative
    g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6)

    i really dont know how to do this, i tried but got nowhere.
    I'm not sure... Are you asked to differentiate g by using the product rule ?


    Look at this :

    g(x)=x \cdot f(x), right ?

    What does the product rule says ?
    (uv)'=u'v+uv'

    Where is the product here ?
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  9. #9
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    it is asking to show that g(x) =xe^(x(4-x)/6) has the derivative
    g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6) by using the previous answer and by using the product rule where the product rule is:

    k'(x) = f'(x) g(x) + f(x) g'(x)
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  10. #10
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    Quote Originally Posted by Sweeties View Post
    it is asking to show that g(x) =xe^(x(4-x)/6) has the derivative
    g'(x)=1/3 (3+2x-x^2)e^(x(x-4)/6) by using the previous answer and by using the product rule where the product rule is:

    k'(x) = f'(x) g(x) + f(x) g'(x)
    Ok then, we know that g(x)=x*f(x), because f(x)=e^(x(4-x)/6)

    --> g'(x)=(x)'*f(x)+x*f '(x)=f(x)+x*f '(x)

    Do you agree ?
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  11. #11
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    ok. get this part:

    g(x) = x.f(x) = xe^(x(4-x)/6)

    so

    g'(x) = x'.f(x) + x.f'(x)

    giving

    e^(x(4-x)/6) + x . e^(x(4-x)/6) (2/3 - 1/3x) =

    e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

    but I don't see how this will get to
    =1/3 (3+2x-x^2)e^(x(x-4)/6)
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  12. #12
    Moo
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    Quote Originally Posted by Sweeties View Post
    e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)
    Yes !

    Now, observe :

    e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

    =e^(x(4-x)/6) (1+x(1/3(2-x)))

    =e^(x(4-x)/6) * (1+2/3 x - x)

    =...
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  13. #13
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    Quote Originally Posted by Moo View Post
    Yes !

    Now, observe :

    e^(x(4-x)/6) + x . e^(x(4-x)/6) . 1/3(2-x)

    =e^(x(4-x)/6) (1+x(1/3(2-x)))
    sorry, i cant see how you went from the first to the second line
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  14. #14
    Moo
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    Quote Originally Posted by Sweeties View Post
    sorry, i cant see how you went from the first to the second line
    I factorised
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  15. #15
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    aha! so you did.
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