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Math Help - Solve 1

  1. #1
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    Post Solve 1

    How would I solve these problems using a short cut formula, simplify results, collecting like terms if possible. I am totally lost with derivative, quotient and product rule.

    f(x) =-8

    g(x)=4x^-5+20x^-1

    k(x)=15x 2/3

    r(x)=16x^7-8x^5+4x^3-2x+1

    f(x) = (√x+1)(Vx-1)
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  2. #2
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    Ok. Some basic rules. The derivative of a constant is zero. That could be 1,2,3, pi, e, phi, etc. - the derivatives of all of these are zero.

    If you have a polynomial in the form of y=x^n, then y'=nx^{n-1} So you can apply both of these rules to your f(x),g(x), and r(x).

    On k(x), is that supposed to be 15x^{\frac{2}{3}}, or 15x*\frac{2}{3}? I can't tell. And on the second f(x) term, is that supposed to be \frac{\sqrt{x+1}}{\sqrt{x-1}}?
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  3. #3
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    Quote Originally Posted by Hallah_az
    How would I solve these problems using a short cut formula, simplify results, collecting like terms if possible. I am totally lost with derivative, quotient and product rule.

    f(x) =-8

    g(x)=4x^-5+20x^-1

    k(x)=15x 2/3

    r(x)=16x^7-8x^5+4x^3-2x+1

    f(x) = (√x+1)(Vx-1)
    f'(x)=(-8)'=0

    g'(x)=(4x^{-5}+20x^{-1})'=-20x^{-6}-20x^{-2}

    k'(x)=(15x^{2/3})'=15\cdot\frac{2}{3}x^{-1/3}=10x^{-1/3}

    r'(x)=(16x^7-8x^5+4x^3-2x+1)'= 112x^6-40x^4+12x^2-2

    f'(x)=[(\sqrt{x}+1)(\sqrt{x}-1)]'=[x-1]'=1

    Next time place the thread in the appropraite section.
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