I am trying to find the derivative of $2\sqrt{x-3}$ using the limit definition of a derivative.

What I did is $$\lim_{h \to 0} \frac {2 \sqrt {(x+h)-3} -2 \sqrt {x-3)})}{h}$$

I multiplied the denominator and numerator by $$2 \sqrt {(x+h)-3}+ 2 \sqrt {x-3}$$

After simplifying I got $$\frac {4}{ 2 \sqrt {(x+h)-3}+2 \sqrt {x-3}}$$

But I am having problem finding the correct answer.

  • 1
    $\begingroup$ After 330 reputation you should have learned to use proper formatting by now.. $\endgroup$ – nbubis Jan 31 '13 at 19:37
  • $\begingroup$ just visit and learn mathjax meta.math.stackexchange.com/questions/5020/… $\endgroup$ – Argha Jan 31 '13 at 19:44
  • $\begingroup$ How would I rationalize the numerator is what I am wondering because at the end of my problem I got ((4))/((4 squareroot(x)-6)) $\endgroup$ – Fernando Martinez Jan 31 '13 at 19:51
  • $\begingroup$ @FernandoMartinez: Again, please take a look at the link provided in Argha's comment. It isn't too difficult to format things using MathJax. $\endgroup$ – Thomas Jan 31 '13 at 19:53
  • $\begingroup$ Ok I will try it but I am not sure what I am doing wrong in this problem $\endgroup$ – Fernando Martinez Jan 31 '13 at 19:59

$\cfrac{(2\sqrt{x+h}-3)-(2\sqrt{x}-3)}{h}=2\cfrac{\sqrt{x+h}-\sqrt{x}}{h}=2\cfrac{(\sqrt{x+h}-\sqrt{x})(\sqrt{x+h}+\sqrt{x})}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{(x+h)-(x)}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{h}{h(\sqrt{x+h}+\sqrt{x})}=2\cfrac{1}{\sqrt{x+h}+\sqrt{x}}\xrightarrow{\scriptscriptstyle h\to0} 2\cfrac{1}{\sqrt{x}+\sqrt{x}}=\cfrac{1}{\sqrt{x}}$

  • 1
    $\begingroup$ You misplaced the 2 after the second $=$. $\endgroup$ – Rick Decker Jan 31 '13 at 19:54
  • $\begingroup$ @RickDecker : Fixed. Ty :) $\endgroup$ – xavierm02 Jan 31 '13 at 20:11
  • $\begingroup$ I have a question what if you chose to multiply the top and bottom by 2 square root(x+h)-3 + 2 square root(x)-3 would that be incorrect. $\endgroup$ – Fernando Martinez Jan 31 '13 at 20:16
  • $\begingroup$ @FernandoMartinez : No. But you make it more complicated for no reason by keeping the $3$s and if you remove then $3$s the keeping the $2$s is useless since you would simplify it right away. $\endgroup$ – xavierm02 Jan 31 '13 at 20:25
  • $\begingroup$ @xaviero2 I see ty. $\endgroup$ – Fernando Martinez Jan 31 '13 at 20:26

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.