# Using the Quotient Rule with the Chain Rule

1. Nov 5, 2006

### TooManyHours

Hi,

I'm struggling to understand how to find the derivative of something like this...

[(2x - 1)^2] / [(x - 2)^3]

The answer in my book says it is supposed to be [-(2x - 1)(2x + 5)] / [(x - 2)^4]

How do I use the chain rule with the quotient rule at the same time?

I even multiplied it all out and tried to find the derivative with just the quotient rule but I can't get the book's final simplifed value.

Thanks

2. Nov 5, 2006

### loto

Remember, that something like 1 / x is equal to x^-1. You can put everything in the numerator, and then use the product rule first.

3. Nov 5, 2006

I wrote that a million times before, but I'm too lazy to search. So, let f(x) = (2x -1)^2, g(x) = (x - 2)^3, and the original function h(x) = f(x) / g(x) You have:

h ' (x) = ( f ' (x)*g(x) - f(x)*g ' (x) ) / ( g(x)^2 ).

4. Nov 5, 2006

### TooManyHours

I know that's the quotient rule. My problem is that when I apply it to my problem, I can't seem to get the simplified answer from the book.

For example, if the denominator is (x - 2)^3 and it's squared, shouldn't that produce (x - 2)^6. Therefore, how could the answer have a denominator of only (x -2 )^4?

Thanks

5. Nov 5, 2006

### Hurkyl

Staff Emeritus
Make an equation out of it.

$$\frac{M}{(x-2)^6} = \frac{N}{(x-2)^4}$$

Can this ever be true? When?

6. Nov 6, 2006

### HallsofIvy

Staff Emeritus
Because you are going to have a factor of at least (x-2)^2 in each term in the numerator and that cancels.

7. Nov 6, 2006

### TooManyHours

Thanks a lot. Once I factored out (x-2)^2 from the numerator and denominator, I was able to reduce my answer to what was shown in the book...