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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.

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

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    $\begingroup$ 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

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