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

  • 11
    $\begingroup$ Nice handwriting. $\endgroup$ – Daniel W. Farlow Jan 17 '15 at 19:57
  • 2
    $\begingroup$ You changed a sum to a product in the third line from the bottom. $\endgroup$ – symplectomorphic Jan 17 '15 at 20:04
  • 3
    $\begingroup$ @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
  • 3
    $\begingroup$ I wish I could draw lines that straight. Damn. $\endgroup$ – Emily Jan 17 '15 at 20:21
  • 4
    $\begingroup$ @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

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*}

  • $\begingroup$ @Adrinao I see where my error is now! Thanks! $\endgroup$ – George Jan 17 '15 at 20:10

The error you made was when you passed from :

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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.