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So I'm trying to find the derivative of $f(x) = 3x + 4\sqrt{x}$

So far I got this far:

1.

$\displaystyle \frac{3(a+h)+ 4\sqrt{a+h} - (3a + 4\sqrt{a})}{h}$

2.

$\displaystyle \frac{3a + 3h + 4\sqrt{a+h} - (3a + 4\sqrt{a})}{h}$

3.

$\displaystyle \frac{3h + 4\sqrt{a+h} - 4\sqrt{a}}{h}$

I'm not sure what to do after step three in solving the square roots.

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    $\begingroup$ Do you have to use the limit definition? Have you already covered the sum and constant multiple rules for derivatives, and the derivatives of the identity and square root functions? $\endgroup$ – David Mitra Feb 2 '14 at 20:17
  • $\begingroup$ Otherwise, break your expression up as $3+4\cdot{\sqrt{a+h}-\sqrt a\over h} $. Then use the procedure of the answers below to deal with the square root term. $\endgroup$ – David Mitra Feb 2 '14 at 20:20
  • $\begingroup$ It would be a good idea if you edited your question and clarified that you have to find the derivative from first principles. $\endgroup$ – Geoff Pointer Feb 2 '14 at 20:33
  • $\begingroup$ @DavidMitra You should probably post this as an answer. The sum rule and the constant product rule are exactly what he needs. I would post it as an answer myself but that would just be rude. $\endgroup$ – Chris Feb 3 '14 at 16:46
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If you've heard of rationalizing the denominator, then now you get to rationalize the numerator; multiply top and bottom by $4(\sqrt{a+h}+\sqrt{a})$; the top will now be free of square roots. The denominator won't be, but you will be able to take a limit. (It would probably be easier to separate the $3h/h$ out first.)

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Use the fact that

$$\frac{\sqrt{a+h}-\sqrt{a}}{h} = \frac{(\sqrt{a+h}-\sqrt{a})(\sqrt{a+h}+\sqrt{a})}{h(\sqrt{a+h}+\sqrt{a})}= \frac{1}{\sqrt{a+h}+\sqrt{a}} \to \frac{1}{2\sqrt{a}}$$

as $h\to 0$.

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