# Thread: Easy Derivative Problem that I don't Understand

1. ## Easy Derivative Problem that I don't Understand

I'm not sure how to attack this. I've used online Derivative Calculators and they only help me to go from point A to B quickly. The only problem is that I don't know what route it takes to arrive at point B.

So far I've just guess at a few of the answers as you can see.

Here's the problem:

2. Why are you using an online calculator? What class is this that they haven't taught you how to take the derivative, but expect you to know the definition of it?

3. It's Calculus I

They have taught me how to take the derivative using the difference quotient. Last class just started talking about the short cuts.

It's just the fractions, and the over all question I'm not sure about.

4. The difference quotient, and what is quote in the image are two different things.

The difference quotient is quite simply:

$\displaystyle f(x+h) = \frac{f(x+h)-f(x)}{h}$.

You should recognize that from Pre-Cal. With our function $\displaystyle f(x) = \frac{6}{x-3}$:

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{ \frac{6}{x+h-3}-\frac{6}{x-3}}{h}\Rightarrow$

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{ \frac{(x-3)6}{x+h-3}-\frac{(x+h-3)6}{x-3}}{h}\Rightarrow$

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{\frac{6x-18-6x-6h+18}{(x+h-3)(x-3)}}{h} \Rightarrow$

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{-6h}{h(x+h-3)(x-3)} \Rightarrow$

Here we can cancel an h from the top and bottom and we are left with:

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{-6}{(x+h-3)(x-3)} \Rightarrow$

Now the limit as "h" approaches zero is just:

$\displaystyle \lim_{h\to 0} f(x+h) =\frac{-6}{(x+-3)^{2}}$

And you are done. These problems look a lot more difficult than they are. The more you do these problems, the more you will realize that the h in the denominator will ALWAYS cancel out with your h's in the numerator. If you have an "h" in parentheses like we did here, do not worry about it, since that individual term will go to zero.

5. While ANDS!'s solution is certainly correct, I want to point out that there are two common variations of the definition of a derivative. One is where the limit variable (usually h) goes to zero:

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$

Another form of this is where you have a limit variable approaching x instead of zero:

$\displaystyle f'(x) = \lim_{t \to x} \frac {f(x)-f(t)}{x-t}$

Both versions give the same result. It appears that you should be using the second version for your assignment, since it indicates the limit as $\displaystyle t \to x$ instead of $\displaystyle h \to 0$.

To get you started, plugging the function in gives something like this:

$\displaystyle f'(x) = \lim_{t \to x} \frac {\frac{6}{x-3}-\frac{6}{t-3}}{x-t}$

6. Originally Posted by drumist

To get you started, plugging the function in gives something like this:

$\displaystyle f'(x) = \lim_{t \to x} \frac {\frac{6}{x-3}-\frac{6}{t-3}}{x-t}$

After that I get,

$\displaystyle f'(x)= \lim_{t\to x} \frac {\frac {6t-6x-6}{xt-3x-3t+9}}{x-t}$

And I figure next to simplify:

( I've already distributed the $\displaystyle \frac {2}{3}$
which came from $\displaystyle \frac {6}{9}$ )

$\displaystyle f'(x)= \lim_{t\to x} \frac {\frac {2xt+4x-4t}{3}}{x-t}$

I don't know if that's right.