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I am trying to use the Derivative Function Definition to show that $f(x) = x^{-1/2}$, where $f'(x)=-\frac{1}{2}x^{-3/2}$. I tried 3 different times, but I think there are some algebra mistakes.

I was thinking if I can cancel out the $\Delta x$ from the denominator, I can then apply $\lim_{x\to 0}\Delta x$ and get my answer. But I only gotten $-\frac{1}{2}x$ so far. My question is, is my approach incorrect? Or there is algebra mistake I kept repeating?

PS: I learned about the power rule, and derivative rules last week. But I would really like to learn where is my mistake or error.

My work

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    $\begingroup$ Nice handwriting. $\endgroup$ – Daniel W. Farlow Jan 17 '15 at 19:57
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    $\begingroup$ You changed a sum to a product in the third line from the bottom. $\endgroup$ – symplectomorphic Jan 17 '15 at 20:04
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    $\begingroup$ @induktio the first thing I thought when I was reading this was "man, I sure wish my work was that neat!" $\endgroup$ – DanZimm Jan 17 '15 at 20:09
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    $\begingroup$ I wish I could draw lines that straight. Damn. $\endgroup$ – Emily Jan 17 '15 at 20:21
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    $\begingroup$ @DanZimm Right? I just wish more people knew the beauty and power of LaTeX...my handwriting is decent, but TeX...nothing can compare with the great Knuth! $\endgroup$ – Daniel W. Farlow Jan 17 '15 at 20:35
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It's very good up to near the end. Your addition suddenly turned into multiplication. You should instead have: \begin{align*} f'(x) &= \cdots \\ &= \lim_{\Delta x \to 0} \frac{-\Delta x}{(\Delta x)(\sqrt{x + \Delta x})(\sqrt x) \cdot [\sqrt x + \sqrt{x + \Delta x}]} \\ &= \lim_{\Delta x \to 0} \frac{-1}{(\sqrt{x + \Delta x})(\sqrt x) \cdot [\sqrt x + \sqrt{x + \Delta x}]} \\ &= \frac{-1}{(\sqrt{x})(\sqrt x) \cdot [\sqrt x + \sqrt{x}]} \\ &= \frac{-1}{x \cdot [2\sqrt x]} \\ &=\frac{-1}{2}x^{-3/2} \end{align*}

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  • $\begingroup$ @Adrinao I see where my error is now! Thanks! $\endgroup$ – George Jan 17 '15 at 20:10
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The error you made was when you passed from :

$$ \lim_{\Delta x\to0}\dfrac{(\Delta x)(-1)}{(\Delta x)\left(\sqrt{x+\Delta x}\right)\left(\sqrt{x}\right)\left[\sqrt{x}\color{#C00}{\boldsymbol+}\left(\sqrt{x+\Delta x}\right)\right]}, $$

to

$$\lim_{\Delta x\to0}(-1)\dfrac{1}{\left(\sqrt{x+\Delta x}\right)\left(\sqrt{x}\right)\left[\sqrt{x}\color{#C00}{\small\bullet}\left(\sqrt{x+\Delta x}\right)\right]}.$$

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