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How do you find the derivative of $\sqrt{x}$ using the symmetric derivative formula?

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}. $$

I got stuck on trying to remove the h from the denominator.

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closed as off-topic by Xander Henderson, user99914, Did, Namaste, Nils Matthes Jul 13 '18 at 13:37

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  • $\begingroup$ You can find it here math.stackexchange.com/a/164713/568718 $\endgroup$ – tien lee Jul 13 '18 at 3:55
  • $\begingroup$ I am unfamiliar with the "symmetric derivative formula." Perhaps you could improve your question by adding a definition, and letting us know what you have tried and where you are stuck? $\endgroup$ – Xander Henderson Jul 13 '18 at 3:59
  • $\begingroup$ @tienlee the link is one-sided, OP needs two-sided $\endgroup$ – gt6989b Jul 13 '18 at 4:00
  • $\begingroup$ $\sqrt{x}$ is differentiable, so the symmetric derivative is the same as the ordinary derivative. I don't think I understand what you're asking. $\endgroup$ – saulspatz Jul 13 '18 at 4:01
  • $\begingroup$ @tienlee en.wikipedia.org/wiki/Symmetric_derivative $\endgroup$ – saulspatz Jul 13 '18 at 4:02
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I am assuming the symmetric derivative formula (also known as the symmetric difference) means that $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}. $$ Note that in your case you have to look at $$ \begin{split} L(h) &= \frac{\sqrt{x+h} - \sqrt{x-h}}{2h} \\ &= \frac{\sqrt{x+h} - \sqrt{x-h}}{2h} \times \frac{\sqrt{x+h} + \sqrt{x-h}}{\sqrt{x+h} + \sqrt{x-h}} \\ &= \frac{(x+h) - (x-h)} {2h \left(\sqrt{x+h} + \sqrt{x-h}\right)} \end{split} $$ Can you take it from here?

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  • $\begingroup$ Thanks for the help. I'd just like clarification on the notation: L(h) $\endgroup$ – Nicolas Jul 13 '18 at 4:12
  • $\begingroup$ @Nicolas $L(h)$ is just some name for a function depending on $h$, for convenient notation... $\endgroup$ – gt6989b Jul 13 '18 at 4:18

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