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Math Help - Help finding derivative

  1. #1
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    Help finding derivative

    Find f''(x) if f(x)=\frac{x^2}{\sqrt{x^2+4}}

    so which method would be easier to solve for? The quotient rule or should I bring \sqrt{x^2+4} to the numerator as (x^2+4)^-^\frac{1}{2} and use the product rule and chain rule?
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  2. #2
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    Quote Originally Posted by ascendancy523 View Post
    Find f''(x) if f(x)=\frac{x^2}{\sqrt{x^2+4}}

    so which method would be easier to solve for? The quotient rule or should I bring \sqrt{x^2+4} to the numerator as (x^2+4)^-^\frac{1}{2} and use the product rule and chain rule?
    Either way will involve using the chain rule. The product rule is always easier though, because you don't have to worry about the order.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Either way will involve using the chain rule. The product rule is always easier though, because you don't have to worry about the order.
    So then

    f'(x)=x^2(-\frac{1}{2})(x^2+4)^-^\frac{3}{2}(2x)+(x^2+4)^-^\frac{1}{2}(2x)

    =(-x^3)(x^2+4)^-^\frac{3}{2}+(x^2+4)^-^\frac{1}{2}(2x)

    then I can't remember if it's acceptable to distribute, so
    =(-x^5-4x^3)^-^\frac{3}{2}+(2x^3+8x)^-^\frac{1}{2}
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  4. #4
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    Quote Originally Posted by ascendancy523 View Post
    So then

    f'(x)=x^2(-\frac{1}{2})(x^2+4)^-^\frac{3}{2}(2x)+(x^2+4)^-^\frac{1}{2}(2x)

    =(-x^3)(x^2+4)^-^\frac{3}{2}+(x^2+4)^-^\frac{1}{2}(2x)


    then I can't remember if it's acceptable to distribute, so
    =(-x^5-4x^3)^-^\frac{3}{2}+(2x^3+8x)^-^\frac{1}{2}
    You can't distribute because what is in parenthesis is raised to a power. If you want to distribute, you'd need to raise the binomial to the exponent first (PEMDAS).

    Edit: However, you can factor out (x^2+4)^-^\frac{3}{2}
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  5. #5
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    Quote Originally Posted by IKoRuPT View Post
    You can't distribute because what is in parenthesis is raised to a power. If you want to distribute, you'd need to raise the binomial to the exponent first (PEMDAS).

    Edit: However, you can factor out (x^2+4)^-^\frac{3}{2}
    So then something like this...
    (x^2+4)^-^\frac{3}{2}[(-x^3)+(x^2+4)(2x)]??
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  6. #6
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    You can't take out (x^2 + 4)^{-\frac{3}{2}}, but you can take out (x^2 + 4)^{-\frac{1}{2}}.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    You can't take out (x^2 + 4)^{-\frac{3}{2}}, but you can take out (x^2 + 4)^{-\frac{1}{2}}.
    ah, that's right..so maybe

    (x^2+4)^-^\frac{1}{2}[(-x^3)(x^2+4)^-^1+(2x)]??
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