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

Question

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.

Answer 1

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?