Finding the derivative $f(x)=\sqrt{x^2 -9}$, I need to find the slope at a=5, using the definition for the function $f(x)=\sqrt{x^2 -9}$,
$$f'(x) = \lim_{\Delta x \to 0} {f(x+\Delta x)\over \Delta x}$$
The answer book says the slope is ${1\over 4}$
Here's what I did,
$$f'(x) = \lim_{\Delta x \to 0} {(\sqrt{(x+\Delta x)^2 -9} - \sqrt {x^2 -9} )(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)})\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(1)\\=\lim_{\Delta x \to 0} {(x+\Delta x)^2 -9 -x^2 +9\over \Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(2)\\=\lim_{\Delta x \to 0}{x^2 +2x \Delta x+ \Delta x^2 -x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(3)\\=\lim_{\Delta x \to 0}{2x\Delta x +\Delta x^2\over\Delta x(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(4)\\=\lim_{\Delta x \to 0} {2x+\Delta x\over(\sqrt{(x+\Delta x)^2 -9}+ \sqrt{x^2-9)}}
(5)\\={2x\over \sqrt{x^2-9+x^2-9}}
(6)\\={2x\over2 \sqrt{x^2 -9}}
(7)\\={x\over \sqrt {x^2 -9}} (8)$$
Now I substitute 5, and I don't get 1/4!!
What have I done wrong??
Thanks

(5)to (6) step  !!!you have error $$\sqrt{(x+\Delta x)^2-9}+\sqrt{(x)^2-9} \neq \sqrt{(x)^2-9+(x)^2-9} $$in fact  $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$
