If anyone can help with any of these questions please post
1) y= √(16-x) , where y= 5
2) y= √(x-7) , at x= 16
3) y= 8/(√(x+11)) , at x=5
Ah, the long way. All right, I'll do the first then you can try the other two. They follow the same pattern.
$\displaystyle y^{\prime}(x) = \lim_{h \to 0} \frac{y(x + h) - y(x)}{h}$
$\displaystyle = \lim_{h \to 0} \frac{\sqrt{16 - (x + h)} - \sqrt{16 - x}}{h}$
This next step will likely go against every simplification rule you ever learned in Algebra. Then again, this is Calculus, not Algebra.
We're going to rationalize the numerator. Why? Because it works.
$\displaystyle = \lim_{h \to 0} \frac{\sqrt{16 - (x + h)} - \sqrt{16 - x}}{h} \cdot \frac{\sqrt{16 - (x + h)} + \sqrt{16 - x}}{\sqrt{16 - (x + h)} + \sqrt{16 - x}}$
$\displaystyle = \lim_{h \to 0} \frac{(16 - (x + h)) - (16 - x)}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$
$\displaystyle = \lim_{h \to 0} \frac{16 - x - h -16 + x}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$
$\displaystyle = \lim_{h \to 0} \frac{-h}{h(\sqrt{16 - (x + h)} + \sqrt{16 - x})}$
Now divide:
$\displaystyle = \lim_{h \to 0} \frac{-1}{\sqrt{16 - (x + h)} + \sqrt{16 - x}}$
Now take the limit:
$\displaystyle = \frac{-1}{\sqrt{16 - x} + \sqrt{16 - x}}$
$\displaystyle = \frac{-1}{2\sqrt{16 - x}}$
Now go ahead and put the x value in.
-Dan