# Using Derivative Function Definition to find the Derivative

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.

• Nice handwriting. – Daniel W. Farlow Jan 17 '15 at 19:57
• You changed a sum to a product in the third line from the bottom. – symplectomorphic Jan 17 '15 at 20:04
• @induktio the first thing I thought when I was reading this was "man, I sure wish my work was that neat!" – DanZimm Jan 17 '15 at 20:09
• I wish I could draw lines that straight. Damn. – Emily Jan 17 '15 at 20:21
• @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! – Daniel W. Farlow Jan 17 '15 at 20:35

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