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Math Help - Problem using the difference quotient

  1. #1
    Junior Member astartleddeer's Avatar
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    Problem using the difference quotient

    Hello,


    I'm trying to differentiate a sum of functions by means of the difference quotient from a question in my book. The derivative is a given, but I can't end up at the final result (I can, but only if I use  anx^{n -1} )

    I feel like I'm close but I now need assistance.

     f(x) = \frac{7}{2} x^\frac{1}{2} + \frac{3}{x^2} - 9

     f(x + \Delta x) = \frac{7}{2} (x + \Delta x)^\frac{1}{2} + \frac{3}{(x + \Delta x)^2} - 9

     f(x + \Delta x) - f(x) = \left[\frac{7}{2} (x + \Delta x)^\frac{1}{2} + \frac{3}{(x + \Delta x)^2} - 9\right] - \left[\frac{7}{2} x^\frac{1}{2} + \frac{3}{x^2} - 9\right]

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

    Now this is where I'm stuck. If I assume I can eliminate  \left[(x +\Delta x)^\frac{1}{2} - x^\frac{1}{2}\right] by way of the difference of two squares. Then I can multiply both the numerator and denominator by  \left[(x +\Delta x)^\frac{1}{2} + x^\frac{1}{2}\right] in order to replace it with  (x + \Delta x) - x in the numerator.

    As for  \left[\frac{1}{(x + \Delta x)^2} - \frac{1}{x^2}\right] again I think I can eliminate the squares to a more easier form to deal with by way of the difference of two squares but don't know how?

    Thank you for your attention.



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  2. #2
    MHF Contributor Reckoner's Avatar
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    Re: Problem using the difference quotient

    Quote Originally Posted by astartleddeer View Post
     \frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\frac{7}{2}\left[(x +\Delta x)^\frac{1}{2} - x^\frac{1}{2}\right] + 3\left[ \frac{1}{(x + \Delta x)^2} - \frac{1}{x^2}\right]}{\Delta x}
    On the left side, rationalize the numerator (this is a strategy that is often helpful in situations like this), and on the right, combine fractions and expand.

    \lim_{\Delta x\to0}\frac{f(x+\Delta x) - f(x)}{\Delta x} = \lim_{\Delta x\to0}\frac{7(x+\Delta x)^{1/2}-7x^{1/2}}{2\Delta x} + \lim_{\Delta x\to0}\frac1{\Delta x}\left[\frac3{(x+\Delta x)^2} - \frac3{x^2}\right]

    = \frac72\lim_{\Delta x\to0}\frac{(x+\Delta x)^{1/2}-x^{1/2}}{\Delta x} + 3\lim_{\Delta x\to0}\frac{x^2 - (x+\Delta x)^2}{x^2\Delta x(x+\Delta x)^2}

    = \frac72\lim_{\Delta x\to0}\frac{\left[(x+\Delta x)^{1/2}-x^{1/2}\right]\left[(x+\Delta x)^{1/2}+x^{1/2}\right]}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} + 3\lim_{\Delta x\to0}\frac{x^2 - \left(x^2+2x\Delta x+\Delta x^2\right)}{x^2\Delta x(x+\Delta x)^2}

    = \frac72\lim_{\Delta x\to0}\frac{(x+\Delta x)-x}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} - 3\lim_{\Delta x\to0}\frac{2x\Delta x+\Delta x^2}{x^2\Delta x(x+\Delta x)^2}

    = \frac72\lim_{\Delta x\to0}\frac{\Delta x}{\Delta x\left[(x+\Delta x)^{1/2}+x^{1/2}\right]} - 3\lim_{\Delta x\to0}\frac{\Delta x\left(2x+\Delta x\right)}{x^2\Delta x(x+\Delta x)^2}

    = \frac72\lim_{\Delta x\to0}\frac1{(x+\Delta x)^{1/2}+x^{1/2}} - 3\lim_{\Delta x\to0}\frac{2x+\Delta x}{x^2(x+\Delta x)^2}

    = \frac7{2\left(x^{1/2}+x^{1/2}\right)} - \frac{3(2x)}{x^2\cdot x^2}

    = \frac7{4\sqrt x} - \frac6{x^3}
    Thanks from astartleddeer
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  3. #3
    Junior Member astartleddeer's Avatar
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    Re: Problem using the difference quotient

    Beautiful! Thanks!
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