# 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

• You have asked several similar questions already. Let's take a break. In every question, it's because you made a simply arithmetic error. Perhaps explore one answer and work your other problems more carefully before asking yet another almost identical question. Jan 9, 2015 at 16:58
• Your calculation was fine, and then there were a couple of algebra errors in a row. As a minor suggestion, since we are interested in $x=5$, it might have been a good idea not to proceed "generally," but instead to use $x=5$ from the beginning. Jan 9, 2015 at 17:00
• @Arkamis I don't see a problem with the user asking the questions he/she did. If it bothers you that their mistakes are arithmetic, maybe you should move on. Unless the user is violating StackExchange rules, it isn't for any of us to tell them what questions they can and can't ask. Jan 9, 2015 at 17:02
• @user46944 It is generally frowned-upon on this site to rapidly and serially ask questions that are abstract duplicates. Jan 9, 2015 at 17:03
• (Cont.) So there were errors. But in fact the slope is not $\frac{1}{4}$, so there is also an error in the answer file. Jan 9, 2015 at 17:05

(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 $$\sqrt{a+b} \neq \sqrt{a} +\sqrt{b}$$
In step 6, the denominator should be $2 \sqrt{x^2-9}$.
• Also, are you sure the book's answer is $\frac{1}{4}$? It should be $\frac{5}{4}$. Jan 9, 2015 at 16:58
• $\sqrt{x^2-9}+\sqrt{x^2-9}=2\sqrt{x^2-9}$. Jan 9, 2015 at 17:01