# “Find $f '(a)$” explaining in english

Find $f '(a)$, $f(x) = \frac{x^2 + 1}{x - 5}$

$f'(a)=$ [answer box here]

I understand that $f'(a)$ is the derivative, but i don't understand what the derivative is. Or how to solve this. I am used to finding a tangent slope with two points. This one doesn't have that. So i don't understand how i can use the definition to solve this. Thanks,

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Are you used to finding the tangent slope when the two points get really close? The quotient rule is easier than doing it from scratch (by calculating the limit), but if you don't have the quotient rule handy you can go through all the algebra with the limit. Do you know which method to find the derivative you're expected to use (quotient rule or limit or something else)? –  Mitch Feb 1 '11 at 17:45

HINT: Use the $\frac{f}{g}$ rule which states that $$\Bigl(\frac{f}{g}\Bigr)' = \frac{ g \cdot f' - f \cdot g'}{g^{2}}$$

So here your $f(x)=x^{2}+1$ and $g(x) = x-5$.

So your $$f'(x) = \frac{(x-5) \cdot \frac{d}{dx}\Bigl[x^{2}+1\Bigr] - (x^{2}+1) \cdot \frac{d}{dx}(x-5)}{(x-5)^{2}}$$

After doing this plugin $x=a$.

Okay if you are not comfortable with this we can do this by Product rule. I hope you are familiar with the product rule which states: $$(fg)' = f \cdot g' + g \cdot f'$$

So your $f(x) = x^{2}+1$ and $g(x) = (x-5)^{-1}$. Then by Product rule you have $$(fg)'= (x^{2}+1) \Bigl[ (x-5)^{-2}\Bigr] + \frac{1}{x-5} \Bigl[\frac{d}{dx} (x^{2}+1)\Bigr]$$

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I dont understand any of that. We just started this section. I never seen or heard of a f/g rule. but does that mean.. ((x−5)*(x^2+1)-(x^2+1)*(x−5))/((x−5)^2) –  John Carbonator Feb 1 '11 at 7:26
@john: See the edited answer. –  anonymous Feb 1 '11 at 7:30

In this case I would use the 'quotient rule': $\left( \frac{g}{h} \right)' = \frac{g'h - gh'}{h^2}$.

This means that you have to find the derivative of $g(x) = x^2 +1$ and $h(x) = x -5$

To find $g(x)$ you can use the power rule (a specific case of the chain rule), the sum rule, and the constant rule.

First split $g(x)=x^2 + 1$ into $f(x) =x^2$ and $f(x)=1$.

The power rule is if $f(x) = x^a$ where a is a constant, then $f'(x)=ax^{a-1}$. So looking at $x^2$ you can see that $a = 2$ so $f'(x) = 2x^{2-1} = 2x$, right?

Then you take the constant rule and look at $f(x) = 1$. The constant rule is if $f(x) = a$ where $a$ is some constant, then $f'(x) = 0$. In this case $a=1$ and is constant so $f'(1) = 0$.

Now use the sum rule to put these together. The sum rule is: $(f + g)' = f' + g'$ basically saying that you can take the derivatives separately first and then add them together. So now we take the derivatives that we already calculated and add them: $2x + 0 = 2x = g'(x)$.

If you now take another look at that original quotient rule, you have solved for $g'$... Since you already know $g$ and $h$ (without the $'$) all you need to do is find is $h'$ and $h^2$ and plug them in...

For a good reference, it might be worth it to check out http://en.wikipedia.org/wiki/Derivative.

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The derivative is a limit of the slopes between two points that you're used to finding. As the points get closer together, these slopes between two points approach the slope of the line tangent to the graph of the function at a point. For an excellent explanation of the general ideas here, see this answer of Arturo Magidin.

If you were finding the slope of the line connecting $(a,f(a))$ and a nearby point on the graph, say $(a+\Delta x, f(a+\Delta x))$, where $\Delta x$ is thought of as being small, then you would compute change in $y$ over change in $x$ as $\frac{f(a+\Delta x) - f(a)}{\Delta x}$. The derivative $f'(a)$ is the limit of these quotients as $\Delta x$ goes to $0$. One way to find $f'(a)$ in this case is to first do some algebra with the general expression $\frac{f(a+\Delta x) - f(a)}{\Delta x}$, until you get it in a form where it is clear what happens when $\Delta x$ goes to zero. As indicated in other answers, there are general rules for calculating derivatives of algebraic expressions like this without appeal to limits, but if you haven't learned those yet then the direct method may be the way to go.

My approach to directly computing the limit would be to first break $f$ into simpler functions using polynomial division. This gives $f(x)=26\cdot\frac{1}{x-5}+x+5$. The heaviest algebra will be simplying the combination of $\frac{1}{a+\Delta x - 5}$ and $\frac{1}{a-5}$ in the quotient, but you'll know you're on the right track when you can cancel the $\Delta x$ in the denominator of the whole thing. At this point you should be able to "set $\Delta x = 0$" to compute the limit.

In practice, after first becoming acquainted with the limit definition, everyone learns to use rules for finding derivatives of polynomials and quotients, as indicated in other answers. Those methods are more efficient, but might seem mysterious if you haven't seen derivatives before. An example of using the limit definition is given on the Wikipedia page here.

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