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Math Help - differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

  1. #1
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    differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

    I used the chain rule first for each of the two functions (g(x) and f(x)), getting

    g'(x) = 3/2x^2(x^3 + 2)^-1/2

    f'(x) = -3/2x^2(x^3 + 7)^-3/2

    Then, using the product rule, I get f(x)g'(x) + f'(x)g(x) =

    = [(x^3 +7)^-1/2][3/2x^2(x^3 + 2)^-1/2] + [-3/2x^2(x^3 + 7)^-3/2] [(x^3 + 2)^1/2]

    which is just a mess, and I don't know how (or if it's possible) to simplify it.
    Is that correct so far?
    Ta!
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  2. #2
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    Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

    You can keep going,

    Organise the first 2 brackets into one fraction and do the same for the last two brackets, take the lowest common denominator which should be:

    2(x^3+2)^{\frac{1}{2}}(x^3+7)^{\frac{3}{2}}
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  3. #3
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    Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

    Quote Originally Posted by canger View Post
    I used the chain rule first for each of the two functions (g(x) and f(x)), getting

    g'(x) = 3/2x^2(x^3 + 2)^-1/2

    f'(x) = -3/2x^2(x^3 + 7)^-3/2

    Then, using the product rule, I get f(x)g'(x) + f'(x)g(x) =

    =[(x^3 +7)^-1/2][3/2x^2(x^3 + 2)^-1/2] + [-3/2x^2(x^3 + 7)^-3/2] [(x^3 + 2)^1/2]

    which is just a mess, and I don't know how (or if it's possible) to simplify it.
    Is that correct so far?
    Ta!
    Hi canger,

    Yes it is correct.

    \left[(x^3 +7)^{-\frac{1}{2}}\right]\left[\frac{3x^2}{2}(x^3 + 2)^{-\frac{1}{2}}\right]+\left[-\frac{3x^2}{2}(x^3 + 7)^{-\frac{3}{2}}\right] \left[(x^3 + 2)^\frac{1}{2}\right]

    Since \frac{3x^2}{2} and (x^3 +7)^{-\frac{1}{2}} are common factors you can take it out of the brackets and write separately.

    =\frac{3x^2(x^3 +7)^{-\frac{1}{2}}}{2}\left\{(x^3 + 2)^{-\frac{1}{2}}-(x^3 + 7)^{-1}(x^3 + 2)^\frac{1}{2}\right\}

    =\frac{3x^2(x^3 +7)^{-\frac{1}{2}}}{2}\left\{\frac{1}{(x^3 + 2)^{\frac{1}{2}}}-\frac{(x^3 + 2)^\frac{1}{2}}{(x^3 + 7)}\right\}

    =\frac{3x^2}{2(x^3 +7)^{\frac{1}{2}}}\left\{\frac{1}{(x^3 +  2)^{\frac{1}{2}}}-\frac{(x^3 + 2)^\frac{1}{2}}{(x^3 +  7)}\right\}

    Now take the lowest common denominator of the two fractions inside the brackets and simplify it. Hope you can do it from here.
    Last edited by Sudharaka; July 14th 2011 at 02:35 AM. Reason: Simplification error.
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    Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

    It looks like you just take out the  (x^3 + 7)^-1/2
    Is that right?

    Thank you so much!

    P.S. How did you get your power of (-1/2) look right?
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  5. #5
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    Re: differentiate y= [(x^3 + 2)^1/2][(x^3 +7)^-1/2]

    Quote Originally Posted by canger View Post
    It looks like you just take out the  (x^3 + 7)^-1/2
    Is that right?

    Thank you so much!

    P.S. How did you get your power of (-1/2) look right?
    Yes I didn't notice that. Since (x^3+7)^{-\frac{1}{2}} is also a common factor it could be taken out of the brackets also. I have corrected that in my previous post.

    (x^3+7)^{-\frac{1}{2}}-------["tex"](x^3+7)^{-\frac{1}{2}}["/tex"] I hope now you understand how to write the fractional power correctly.
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