$\displaystyle \frac{1}{h}(\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}})$
$\displaystyle =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)$
$\displaystyle =\frac{1}{h}\left(\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x(x+h)}}\right)\left( \frac{\sqrt{x}+\sqrt{x+h}}{\sqrt{x}+\sqrt{x+h}} \right)$
$\displaystyle =\frac{1}{h}\left(\frac{x -x-h}{\sqrt{x(x+h)}}\right)\left( \frac{1}{\sqrt{x}+\sqrt{x+h}} \right)$
Now continue from there....
I don't know what's going on in your denominator...
$\displaystyle \frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}$
$\displaystyle =\frac{\sqrt{x+1} - \sqrt{a+1}}{x-a}\cdot\frac{\sqrt{x+1} + \sqrt{a+1}}{\sqrt{x+1} + \sqrt{a+1}}$
$\displaystyle =\frac{(x+1) - (a+1)}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}$
$\displaystyle =\frac{x - a}{(x - a)\left(\sqrt{x+1} + \sqrt{a+1}\right)}$
$\displaystyle =\frac1{\sqrt{x+1} + \sqrt{a+1}}$
Hi again,
I am looking through the text book to find how you came to the conclusion to use a square of a difference in this case. the radical is cubed so I would thought to use differences of two cubes?
Or am I confused with the radicals somehow? If I disregard manipulating the numerator and just assume the difference of two cubes will remove the radicals my answer is correct. But I am missing something seriously with my understanding of polynomial manipulation because when I actually do the calculation. I get the result below which shows that something is not correct.
to rationalize the numerator of the fraction ...
$\displaystyle \frac{\sqrt[3]{x} - \sqrt[3]{a}}{x-a}$
note that the denominator can be factored as the difference of two cubes ...
$\displaystyle x - a = (\sqrt[3]{x} - \sqrt[3]{a}) \left[(\sqrt[3]{x})^2 + \sqrt[3]{x} \cdot \sqrt[3]{a} + (\sqrt[3]{a})^2 \right]$
can you finish?
Please can you explain exactly why you came to this conlusion?
I am trying to understand what to look for to solve the problem myself but getting the answer still does not explain the process it takes to reach the conclusion.
Thanks again.