# Finding derivative of $\sqrt{9-x}$

I am trying to find the derivative of $\sqrt{9-x}$ using the definition of a derivative

$$\lim_{h\to 0} \frac {f(a+h)-f(a)}{h}$$

$$\lim_{h\to 0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}$$

So to simplify I multiply by the conjugate

$$\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}$$

which gives me

$$\frac {-2a-h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}$$

I have no idea what to do from here, obviously I can easily get the derivative using other methods but with this one I have no idea how to proceed.

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Check the algebra in your numerator; once you get it right, you'll have a nice cancellation. – Brian M. Scott May 9 '12 at 23:22
1] Your limit should $h \rightarrow 0$ and not $x \rightarrow 0$. 2] $9 - (a + h) - 9 + a = -h$ and not $-2a - h$ – TenaliRaman May 9 '12 at 23:25
@Garmen1778: Please do not make substantive corrections: pointing them out is important, but correction should be left to the person asking the question, to help reinforce the knowledge. – Brian M. Scott May 9 '12 at 23:27
Jordan, you can get the real arrow (as in $h\to 0$) with \to. – Brian M. Scott May 9 '12 at 23:29

## 3 Answers

You made a mistake when doing the multiplication upstairs:

When multiplying $$\Bigl( \color{maroon}{\sqrt{9-(a+h)} }- \color{darkgreen}{\sqrt {9-a}}\ \Bigr)\Bigl(\color{maroon}{\sqrt{9-(a+h)} }+\color{darkgreen}{ \sqrt {9-a}}\ \Bigr),$$ you are using the rule $$(\color{maroon}a-\color{darkgreen}b)(\color{maroon}a+\color{darkgreen}b) =\color{maroon}a^2-\color{darkgreen}b^2$$ So you obtain $$\Bigl(\color{maroon}{\sqrt{9-(a+h)}}\ \Bigr)^2 - \Bigl(\color{darkgreen}{\sqrt {9-a}}\ \Bigr)^2= \bigl(9-(a+h)\bigr) - (9-a) = \color{teal}9\color{purple}{-a}-h\color{teal}{-9}+\color{purple} a= -h.$$

Then, to find your derivative, you have to compute \eqalign{ f'(a)= \lim_{h\rightarrow 0} { -h\over h\bigl( \sqrt{9-(a-h) }+\sqrt{9-a}\ \bigr ) } &=\lim_{h\rightarrow 0} { -1\over \sqrt{9-(a-h) }+\sqrt{9-a} }\cr &= { -1\over \sqrt{9-(a-0) }+\sqrt{9-a} }\cr &= { -1\over \sqrt{9-a }+\sqrt{9-a} }\cr &= { -1\over 2\sqrt{9-a} }. }

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I get that but I am still stuck I am getting $\frac{-1}{\sqrt{9-a} + 3}$ – user138246 May 9 '12 at 23:28
After cancellation, the top is $-1$. The bottom is $\sqrt{9-(a+h)}+\sqrt{9-a}$. As $h$ approaches $0$, the bottom approaches $2\sqrt{9-a}$. – André Nicolas May 9 '12 at 23:31
@AndréNicolas I get it now, I kept mixing up a and h. – user138246 May 9 '12 at 23:36
@Jordan: Good that you have it under control, it is a standard exam question. There is however usually at most one, often $0$ of these. – André Nicolas May 9 '12 at 23:38

Everything you have done is right except for the last step.

\begin{align} &\lim_{h\to0} \frac {\sqrt{9-(a+h)}-\sqrt{9-a}}{h}\cdot \frac{ \sqrt{9-(a+h)}+ \sqrt{9-a}}{\sqrt{9-(a+h)}+\sqrt{9-a}}=\\ &\lim_{h\to0} \frac{9-(a+h)-(9-a)}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\ &\lim_{h\to0} \frac{9-a-h-9+a}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\ &\lim_{h\to0} \frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}=\\ &\lim_{h\to0} \frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}} \end{align}

The limit is then easy to evaluate.

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As other answers well say, you have an error while multiplicating. And also you forgot to put the limit in your last equation. \begin{align} f'(x)&=\lim\limits_{h\to0}\left(-\frac{h}{h(\sqrt{9-(a+h)}+\sqrt{9-a})}\right)\\ &=\lim\limits_{h\to0}\left(-\frac{1}{\sqrt{9-(a+h)}+\sqrt{9-a}}\right)\\ &=-\frac{1}{\sqrt{9-x}+\sqrt{9-x}}\\ &=-\frac{1}{2\sqrt{9-x}} \end{align} But this can be done much easily by this way:

\begin{align} f(x)&=\sqrt{9-x}\\ &=(9-x)^{1/2}\\ f'(x)&=-\frac{1}{2}(9-x)^{-1/2}\\ &=-\frac{1}{2\sqrt{9-x}} \end{align}

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