Finding the derivative $f(x)=\sqrt{x^2 -9}$,

Question

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

Answer 1

In step 6, the denominator should be $2 \sqrt{x^2-9}$.

Answer 2

(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 $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$