Determine the derivative of the function $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ using first principles.
Its been awhile since I've done these.. but I think be first principles it means using the formular $\displaystyle f'(x)= lim (h\rightarrow0) \frac{f(x+h)-f(x)}{h}$
Therefore I have my first step as
$\displaystyle f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h}$
I then rationalized...
$\displaystyle f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h} * \frac{1/{\sqrt{x+h}}+1/{\sqrt{x}}}{1/{\sqrt{x+h}}+1/{\sqrt{x}}}$
And I just can't seem to get a final derivative
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.
$\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$
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...
$\displaystyle \frac{\frac{x-(x+h)}{x(x+h)}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$.
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...
$\displaystyle \frac{-\frac{1}{x^2+xh}}{\frac{1}{\sqrt{h+x}}+\frac{1}{\s qrt{x}}}$
From here there is just a little more simplifying to go and you should get your final answer
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 $\displaystyle \frac{x^a}{x^b}=x^{a-b}$ no matter if a<b or b<a.
That's what I suspected. You need to simplify further.
$\displaystyle -\frac{x^{\frac{1}{2}}}{2x^2}=-\frac{1}{2}(x^{\frac{1}{2}}-x^2)=-\frac{1}{2}(x^{-\frac{3}{2}})$
If you do the normal power rule with $\displaystyle x^{-\frac{1}{2}}$, this is the answer you'll end up with.