# find the derivative of the function using the definition of derivative . state the domain of the function and the domain of its derivative

I'm stumped on yet another assignment problem. I'm not allow to use power rule with this problem so i have to rely on good old $$\frac{f(a+h)-f(a)}{h}$$

so here are the steps ive taken thus far but i cant quite bring it home. 1- $$\lim_{h\to 0}\frac{\frac{1}{\sqrt{t+h}}- \frac{1}{\sqrt{t}}}{h}$$

2- get common denominator $\sqrt{t} \sqrt{t+h}$ $$\lim_{h\to 0}\frac{\frac{\sqrt{t}}{\sqrt{t+h}} - \frac{\sqrt{t+h}}{\sqrt{t}}}{h}$$

3- multiply by conjugate pair $$\lim_{h\to 0}\frac{\frac{\sqrt{t}}{\sqrt{t+h}} - \frac{\sqrt{t+h}}{\sqrt{t}}}{h}* \frac{\sqrt{t}+\sqrt{t+h}}{\sqrt{t}+\sqrt{t+h}}$$

4-multiply across and cancel the h's and i end up with $$\frac{-1}{\sqrt{t+h}\sqrt{t}(\sqrt{t}+\sqrt{t+h} )}$$

this is where im stuck the solutions manual gets to $\frac{-1}{\sqrt{t}\sqrt{t}(\sqrt{t}+\sqrt{t})}$

i have no idea how they could have achieved it? I'm missing an intermediate step can someone please point me in the right direction and i think my algebra is failing me here.

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You're not stuck, you're done! Let $h$ go to $0$, and ... –  trb456 Feb 24 '13 at 16:47
$\sqrt{\cdot}$ is continuous. –  Stefan Feb 24 '13 at 16:47
@ trb456 thank you now i get i!!! i feel like a moron now! thank you for pointing that out to me. –  Miguel Feb 24 '13 at 16:58
Your step (2) is wrong, you don't add the fractions. –  vonbrand Feb 24 '13 at 17:20
Since the square root function is continuous, then you can let the limit "pass through" the radicals, and take $h$ to $0$. We still have to be a bit careful, though. What value(s) of $t$ in the original domain will cause issues in the resulting expression?