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Math Help - derivative of function using accurate notation

  1. #1
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    derivative of function using accurate notation

    hey everyone, so the question is y= -5 over x cubed minus 2 over x + (x+1)

    for -5 over x cubed:using quotient rule, let u= 5 v= x cubed u'= 0 v'= 1

    y'(x)= x cubed(0) - 5(1) all over (x+1)^2

    = 0-5 all over (x+1)^2 ?????

    am I on the right track? or am i supposed to include the minus sign somwhere?
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  2. #2
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    Re: derivative of function using accurate notation

    Hi,

    I do not undestand why you put: v = x cubed and v= 1

    derivative of x cubed = derivative of (x^3) = 3x^2

    Let me know....

    Thanks
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    Re: derivative of function using accurate notation

    or yea, thats right. my bad
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    Re: derivative of function using accurate notation

    my main problem is with the 1st fraction and the negative sign in front of it. what do i do?
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    Re: derivative of function using accurate notation

    Ok,



    Using (u/v)= (uv-uv)/v^2

    and taking u = -5 and v = x^3


    derivative of -5/(x^3) is: [0(x^3)-(-5)3x^2]/(x^3)^2 =

    15x^2/x^6

    = 15/x^4

    Let me know if you need more help
    Thanks from daniels
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    Re: derivative of function using accurate notation

    ok so from the part with [0(x^3)-(-5)3x^2]/(x^3)^2... can u explain how to expand (-5)3x^2 as in the steps please?
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    Re: derivative of function using accurate notation

    oh wait my bad, i meant from the 15x^2/(x^3)^2
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  8. #8
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    Re: derivative of function using accurate notation

    ok (x^3)^2 = (x^3)(x^3) = x^6

    Then: 15x^2/(x^3)^2 = 15x^2/x^6

    If you simplify (divding by x^2)

    The result is 15/x^4
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    Re: derivative of function using accurate notation

    ohhhh....ook thanks thanks thanks (y) was having problems there
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    Re: derivative of function using accurate notation

    so finally, in solving the original question would it be y'= (15/x^4)- (2/x^2) + 2x + 2?
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