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Math Help - Finding the derivative

  1. #1
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    Finding the derivative

    Determine the derivative of the function f(x)=\frac{1}{\sqrt{x}} using first principles.
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  2. #2
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    I don't know what you mean by first principles.

    Think of f(x) as x^{-\frac{1}{2}}. Now use the power rule.
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  3. #3
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    Its been awhile since I've done these.. but I think be first principles it means using the formular f'(x)= lim (h\rightarrow0) \frac{f(x+h)-f(x)}{h}

    Therefore I have my first step as
    f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h}

    I then rationalized...
    f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h} * \frac{1/{\sqrt{x+h}}+1/{\sqrt{x}}}{1/{\sqrt{x+h}}+1/{\sqrt{x}}}

    And I just can't seem to get a final derivative
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  4. #4
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    Oh ok. I see what you meant now. You are using the formula limit definition of a derivative.

    You are on the right track and so close!

    Like you said, you multiply everything by the conjugate of the numerator. Just simplify your last step.

    \frac{\frac{1}{x+h}-\frac{1}{x}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}

    Now let's simplify the numerator again. Find a common denominator for the two terms and combine. The common denominator is x(x+h). So...

    \frac{\frac{x-(x+h)}{x(x+h)}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}.

    Now when you simplify the numerator again, you'll notice you're left with a "-h" that can cancel with the "h" in the denominator leaving...

    \frac{-\frac{1}{x^2+xh}}{\frac{1}{\sqrt{h+x}}+\frac{1}{\s  qrt{x}}}

    From here there is just a little more simplifying to go and you should get your final answer
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  5. #5
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    Ok from the other thread you made I can tell you're still stuck. Try applying the limit now, meaning plug in h=0. From there you'll get a fraction divided by another fraction. Flip the bottom one and multiply. Remember \frac{x^a}{x^b}=x^{a-b} no matter if a<b or b<a.
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  6. #6
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    hmm thank you very much! and my mistake about the other post, I apologize. I think I may have come to final answer. I just wasn't sure before if I was allowed to sub in the h=0 yet.. but I think I ended up getting it. thanks again
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  7. #7
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    No worries. You can show me your final answer and work for the rest of the problem and I'll check it. If you don't want to that's fine too.
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  8. #8
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    f'(x)=\frac{-\sqrt{x}}{2x^2}
    is what i ended up with
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  9. #9
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    That's what I suspected. You need to simplify further.

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

    If you do the normal power rule with x^{-\frac{1}{2}}, this is the answer you'll end up with.
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