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While this particular question is from a calculus class I'm taking, this issue has plagued me for some time; I simply didn't care enough to bother figuring it out. Now, however, it'll cost me if I don't get it, so I'd like to know how to simplify something like this:

$$\frac{\sqrt{2a + 2h + 1} - \sqrt{2a + 1}}{h}$$

This particular problem is attempting to find the derivative of $\sqrt{2x+1}$ (using limits, not just doing the derivative). So, one finds the derivative using limits through the limit as $h \to 0$ of $\frac{f(a+h)-f(a)}{h}$. However, when I put in the function $\sqrt{2x+1}$, I get the above expression. Is there a way to simplify this (and other difficult square roots) or am I just doing this wrong?

Thank you!

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  • $\begingroup$ Use MathJax next time, Ensign. Anyway, the big trick is to rationalize the numerator. $\endgroup$ – Sean Roberson Oct 4 '16 at 17:28
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Observe that using the conjugate of the numerator:

$$\frac{\sqrt{2a + 2h + 1} - \sqrt{2a + 1}}{h}\cdot\frac{\sqrt{2a + 2h + 1} + \sqrt{2a + 1}}{\sqrt{2a+2h+1}+\sqrt{2a+1}}=$$

$$=\frac{2a+2h+1-2a-1}{h\left(\sqrt{2a+2h+1}+\sqrt{2a+1}\right)}=\frac2{\sqrt{2a+2h+1}+\sqrt{2a+1}}\xrightarrow[h\to0]{}\frac2{2\sqrt{2a+1}}=\frac1{\sqrt{2a+1}}$$

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