Can anyone demonstrate the difference quotient f(x) = -6/sqrt(3-4x)
I have worked on that problem for an hour.. tried a few different things.. and I'm flat out lost...
i have gotten so far as -6(sqrt(3-4x-4h) + 6(sqrt(3-4x) / h(sqrt(3-4x))(sqrt(3-4x-4h) however, from there I become unsure what to do. I've thought about multiplying the sqrt on the bottom to give (h)(3+4x)(sqrt(-4h) then multiplying top and bottom by (sqrt(-4h) however, for some reason it seems like doing that is just going to end up leading me in circles.. I thought about multiplying the top using the difference of squares.. but, then that complicates the bottom even more..
If that is what I have to do I will do it.. but, it seems unrealistic to go that route because, I would think I would ultimately complicate the top again clearing the bottom..
you sure didn't pick an easy one ...
$\displaystyle \displaystyle -\frac{6}{h} \left(\frac{1}{\sqrt{3-4(x+h)}} - \frac{1}{\sqrt{3-4x}}\right)$
$\displaystyle \displaystyle -\frac{6}{h} \left(\frac{\sqrt{3-4x}-\sqrt{3-4(x+h)}}{\sqrt{3-4x} \cdot \sqrt{3-4(x+h)}}\right)$
$\displaystyle \displaystyle -\frac{6}{h} \left(\frac{\sqrt{3-4x}-\sqrt{3-4(x+h)}}{\sqrt{3-4x} \cdot \sqrt{3-4(x+h)}}\right) \cdot \left(\frac{\sqrt{3-4x}+\sqrt{3-4(x+h)}}{\sqrt{3-4x}+\sqrt{3-4(x+h)}}\right)$
$\displaystyle \displaystyle -\frac{6}{h} \left(\frac{(3-4x)-[3-4(x+h)]}{\sqrt{3-4x} \cdot \sqrt{3-4(x+h)} \cdot \left[\sqrt{3-4x}+\sqrt{3-4(x+h)}\right]}\right)$
$\displaystyle \displaystyle -\frac{6}{h} \left(\frac{4h}{\sqrt{3-4x} \cdot \sqrt{3-4(x+h)} \cdot \left[\sqrt{3-4x}+\sqrt{3-4(x+h)}\right]}\right)$
$\displaystyle \displaystyle -\frac{24}{\sqrt{3-4x} \cdot \sqrt{3-4(x+h)} \cdot \left[\sqrt{3-4x}+\sqrt{3-4(x+h)}\right]}\right)$
hope that gets you far enough ...