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Math Help - Finding the the derivative of f(x)=(-1/sqrt2x) +2x

  1. #1
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    Finding the the derivative of f(x)=(-1/sqrt2x) +2x

    I am in introductory to derivatives and need to solve this but having a hard time simplifying. I wrote this as the limit as delta x approaches zero of : [(-1/sqrt2(x+deltax)) +2(x+deltax)] - [(-1/2x) +2x] all over delta x.

    I have tried simplify one side of the numerator first, [f(x+deltax)] by finding an LCD and then doing the same for the other side [f(x)]. And then again to combine the two sides of the numerator into one but its a huuuuge mess and I feel there must be an easier way to do things. I don't know what to do at this point. I think if I can get someone to explain to me even the first couple steps in detail I will get the point and be able to finish up. I am aware of multiplying by conjugates to rid of square roots but that is only for 2 terms which doesn't seem to be the case in this question to start out with anyways. Please help...thanks
    Last edited by mr fantastic; July 14th 2011 at 08:51 PM. Reason: Title.
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    Re: Please help me find the the derivative of f(x)=(-1/sqrt2x) +2x

    Quote Originally Posted by jmanna98 View Post
    I am in introductory to derivatives and need to solve this but having a hard time simplifying. I wrote this as the limit as delta x approaches zero of : [(-1/sqrt2(x+deltax)) +2(x+deltax)] - [(-1/2x) +2x] all over delta x.

    I have tried simplify one side of the numerator first, [f(x+deltax)] by finding an LCD and then doing the same for the other side [f(x)]. And then again to combine the two sides of the numerator into one but its a huuuuge mess and I feel there must be an easier way to do things. I don't know what to do at this point. I think if I can get someone to explain to me even the first couple steps in detail I will get the point and be able to finish up. I am aware of multiplying by conjugates to rid of square roots but that is only for 2 terms which doesn't seem to be the case in this question to start out with anyways. Please help...thanks
    to clarify and confirm, is this f(x) ?

    f(x) = 2x - \frac{1}{\sqrt{2x}}
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    Re: Please help me find the the derivative of f(x)=(-1/sqrt2x) +2x

    yes it is
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    Re: Please help me find the the derivative of f(x)=(-1/sqrt2x) +2x

    let h = \Delta x to avoid confusion

    \lim_{h \to 0} \frac{2(x+h) - \frac{1}{\sqrt{2(x+h)}} - \left[2x - \frac{1}{\sqrt{2x}}\right]}{h}

    \lim_{h \to 0} \frac{1}{h}\left[2h - \frac{1}{\sqrt{2(x+h)}} + \frac{1}{\sqrt{2x}}\right]

    \lim_{h \to 0} \frac{1}{h}\left[2h - \frac{\sqrt{2x} - \sqrt{2(x+h)}}{\sqrt{4x(x+h)}}\right]

    \lim_{h \to 0} \frac{1}{h}\left[2h - \frac{\sqrt{2x} - \sqrt{2(x+h)}}{\sqrt{4x(x+h)}} \cdot \frac{\sqrt{2x}+\sqrt{2(x+h)}}{\sqrt{2x}+\sqrt{2(x  +h)}}\right]

    \lim_{h \to 0} \frac{1}{h}\left[2h - \frac{2x - 2(x+h)}{\sqrt{4x(x+h)}\left(\sqrt{2x}+\sqrt{2(x+h)  }\right)}\right]

    \lim_{h \to 0} \frac{1}{h}\left[2h + \frac{2h}{\sqrt{4x(x+h)}\left(\sqrt{2x}+\sqrt{2(x+  h)}\right)}\right]

    \lim_{h \to 0} \frac{h}{h}\left[2 + \frac{2}{\sqrt{4x(x+h)}\left(\sqrt{2x}+\sqrt{2(x+h  )}\right)}\right]

    \lim_{h \to 0} \left[2 + \frac{2}{\sqrt{4x(x+h)}\left(\sqrt{2x}+\sqrt{2(x+h  )}\right)}\right]

    finish it ...
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    Re: Please help me find the the derivative of f(x)=(-1/sqrt2x) +2x

    On the first step I see how you flipped and multiplied the whole equation but what did you do for (2x+h) to change to 2h and where did the 2x go on the other side of the equation? not seeing what step you took to do that?
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    Re: Please help me find the the derivative of f(x)=(-1/sqrt2x) +2x

    Quote Originally Posted by jmanna98 View Post
    On the first step I see how you flipped and multiplied the whole equation but what did you do for (2x+h) to change to 2h and where did the 2x go on the other side of the equation? not seeing what step you took to do that?
    Expand the 2(x + h) and -(2x - 1/....) terms and simplify.
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    Re: Finding the the derivative of f(x)=(-1/sqrt2x) +2x

    wow, of course. silly. That makes much more sense now. ok thank you both. Much appreciated
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