Using the Quotient Rule with the Chain Rule

In summary, the person was struggling to find the derivative of a function using both the chain rule and quotient rule. They were struggling to understand how the denominator of the final simplified answer in their book was different from what they had calculated. They received help and were able to reduce their answer to match the book's answer.
  • #1
TooManyHours
12
0
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
 
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  • #2
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
TooManyHours said:
[(2x - 1)^2] / [(x - 2)^3]

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
radou said:
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 ).

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
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?
Make an equation out of it.

[tex]
\frac{M}{(x-2)^6} = \frac{N}{(x-2)^4}
[/tex]

Can this ever be true? When?
 
  • #6
TooManyHours said:
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
Because you are going to have a factor of at least (x-2)^2 in each term in the numerator and that cancels.
 
  • #7
HallsofIvy said:
Because you are going to have a factor of at least (x-2)^2 in each term in the numerator and that cancels.

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...:smile:
 

Related to Using the Quotient Rule with the Chain Rule

1. What is the Quotient Rule?

The Quotient Rule is a mathematical rule used to find the derivative of a function that is the quotient of two other functions.

2. When is the Quotient Rule used?

The Quotient Rule is used when finding the derivative of a function that is the ratio of two other functions, such as f(x) = g(x)/h(x).

3. What is the Chain Rule?

The Chain Rule is a mathematical rule used to find the derivative of a composite function, where one function is applied to the output of another function.

4. How are the Quotient Rule and Chain Rule related?

The Quotient Rule is often used in conjunction with the Chain Rule when finding the derivative of a function that involves both a quotient and a composite function.

5. Can the Quotient Rule and Chain Rule be used separately?

Yes, the Quotient Rule and Chain Rule can be used separately depending on the type of function being differentiated. However, in some cases, they may need to be used together to find the derivative of a function.

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