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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You can find it here https://math.stackexchange.com/a/164713/568718 – tien lee Jul 13 '18 at 03:55
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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? – Xander Henderson Jul 13 '18 at 03:59
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@tienlee the link is one-sided, OP needs two-sided – gt6989b Jul 13 '18 at 04:00
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$\sqrt{x}$ is differentiable, so the symmetric derivative is the same as the ordinary derivative. I don't think I understand what you're asking. – saulspatz Jul 13 '18 at 04:01
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@tienlee https://en.wikipedia.org/wiki/Symmetric_derivative – saulspatz Jul 13 '18 at 04:02
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You can just add and subtract $\sqrt{x}$ in the numerator and use the known derivative for $\sqrt{x}$ – saulspatz Jul 13 '18 at 04:05
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@saulspatz That assumes that one has already proved that the symmetric derivative reduces to the usual derivative if the function is differentiable. While that is a reasonable approach, one suspects that this is an exercise which was given in order to build computational practice with the symmetric derivative, hence using that result is kind of missing the point. Nicolas: can you explain where, exactly, you got stuck? Can you compute the derivative of $\sqrt{x}$ using the usual definition? – Xander Henderson Jul 13 '18 at 04:10
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@XanderHenderson apparently OP was looking for a transformation to eliminate $h$ from the denominator, as he himself writes in the question – gt6989b Jul 13 '18 at 04:19
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@gt6989b Yes, but the asker is expected to make some effort. As saulspatz indicates, you can just hit this problem over the head with theorems and have an answer. When I first asked for more clarification, the edits (which you corrected) had not yet been made. – Xander Henderson Jul 13 '18 at 04:38
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This question addresses the approach that @saulspatz suggests.. Otherwise, the technique here is almost entirely the same required to compute the derivative of $\sqrt{x}$.. In either case, this question is a duplicate. – Xander Henderson Jul 13 '18 at 04:42
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Possible duplicate of Derivative of square root – amWhy Jul 13 '18 at 11:32
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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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Thanks for the help. I'd just like clarification on the notation: L(h) – Nicolas Jul 13 '18 at 04:12
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@Nicolas $L(h)$ is just some name for a function depending on $h$, for convenient notation... – gt6989b Jul 13 '18 at 04:18