1. ## Finding the derivative

Determine the derivative of the function $f(x)=\frac{1}{\sqrt{x}}$ using first principles.

2. I don't know what you mean by first principles.

Think of f(x) as $x^{-\frac{1}{2}}$. Now use the power rule.

3. Its been awhile since I've done these.. but I think be first principles it means using the formular $f'(x)= lim (h\rightarrow0) \frac{f(x+h)-f(x)}{h}$

Therefore I have my first step as
$f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h}$

I then rationalized...
$f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h} * \frac{1/{\sqrt{x+h}}+1/{\sqrt{x}}}{1/{\sqrt{x+h}}+1/{\sqrt{x}}}$

And I just can't seem to get a final derivative

4. Oh ok. I see what you meant now. You are using the formula limit definition of a derivative.

You are on the right track and so close!

Like you said, you multiply everything by the conjugate of the numerator. Just simplify your last step.

$\frac{\frac{1}{x+h}-\frac{1}{x}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$

Now let's simplify the numerator again. Find a common denominator for the two terms and combine. The common denominator is x(x+h). So...

$\frac{\frac{x-(x+h)}{x(x+h)}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$.

Now when you simplify the numerator again, you'll notice you're left with a "-h" that can cancel with the "h" in the denominator leaving...

$\frac{-\frac{1}{x^2+xh}}{\frac{1}{\sqrt{h+x}}+\frac{1}{\s qrt{x}}}$

From here there is just a little more simplifying to go and you should get your final answer

5. Ok from the other thread you made I can tell you're still stuck. Try applying the limit now, meaning plug in h=0. From there you'll get a fraction divided by another fraction. Flip the bottom one and multiply. Remember $\frac{x^a}{x^b}=x^{a-b}$ no matter if a<b or b<a.

6. hmm thank you very much! and my mistake about the other post, I apologize. I think I may have come to final answer. I just wasn't sure before if I was allowed to sub in the h=0 yet.. but I think I ended up getting it. thanks again

7. No worries. You can show me your final answer and work for the rest of the problem and I'll check it. If you don't want to that's fine too.

8. $f'(x)=\frac{-\sqrt{x}}{2x^2}$
is what i ended up with

9. That's what I suspected. You need to simplify further.

$-\frac{x^{\frac{1}{2}}}{2x^2}=-\frac{1}{2}(x^{\frac{1}{2}}-x^2)=-\frac{1}{2}(x^{-\frac{3}{2}})$

If you do the normal power rule with $x^{-\frac{1}{2}}$, this is the answer you'll end up with.