Determine the derivative (slope of tangent) of x = 6 of the following function using only the methods of first principles only.
$$f(x) = \frac{1}{\sqrt{x-2}}$$
*I am very confused on how to solve this question using first principles. I know that the answer is $\frac{-1}{16}$ but I don't know how to get it. Can anyone please show me how to get to the answer or what I'm doing wrong? Thanks again.
Here's what I did:
Later on, I tried to get the answer but got something similar to it. I tried rationalizing the single expression by itself before putting it together with the whole equation. Am I doing something wrong? Please help!
Best Answer
$$ \frac{1}{\sqrt{4+h}} - \frac{1}{2} = \frac{2-\sqrt{4+h}}{2\sqrt{4+h}} = \frac{(2-\sqrt{4+h})(2+\sqrt{4+h})}{2\sqrt{4+h}(2+\sqrt{4+h})} \\ = \frac{4-(4+h)}{2\sqrt{4+h}(2+\sqrt{4+h})} = \frac{-h}{2\sqrt{4+h}(2+\sqrt{4+h})} $$