Understanding Partial Fractions and the Chain Rule in Integration

In summary, there are two questions being discussed. The first question is about partial fractions and breaking down a complicated denominator. The second question is about the chain rule and how it applies to a composite function. The answer to the first question involves breaking down the denominator into three factors in order to find the right coefficients. The answer to the second question involves using the product rule, not the chain rule, to differentiate the composite function. The correct terminology for this is "composite functions."
  • #1
EvLer
458
0
Hi, I have 2 questions:

1. partial fractions:
if I have following integral: Itegral[(1-2x^2)/(x - x^3)]dx;
my question is do I break down the denominator to x(1-x^2) or do I go further:
x(1-x)(1+x); this way it becomes more complicated;

2. chain rule:
how does chain rule work in this case: f(x)g(x)h(x)?
is it something like f'(x)[g(x)h(x)] + g'(x)[f(x)h(x)] + h'(x)[f(x)g(x)] ? or how would I differentiate this?

Thanks in advance.
 
Physics news on Phys.org
  • #2
1. You need enough unknowns to be able to equate the three coefficients for each power of x in the numerator (1,0,-2). So you'll need to break it up into all three factors. In some cases (not this one) the two factors alone would work, but only if the numerator happens to be just right. It would be like finding the point where two lines intersect in 3D; you can only find it if the lines happen to intersect.

2. Think of it as (fgh)' = f'(gh) + f(gh)'. Your guess is right.
 
  • #3
In first 1-2x^2=1-x^2-x^2 is very helpful. Then you have 2 very easy integrals
 
  • #4
Aaaah, i see. Thanks for help.
 
  • #5
#2 is not the chain rule. That is the product rule. It does work as you have it
 
  • #6
OlderDan said:
#2 is not the chain rule. That is the product rule. It does work as you have it
Right, sorry, mixed up the names :redface:
Chain rule is differentiating what I call "embedded functions".
 
  • #7
EvLer said:
Right, sorry, mixed up the names :redface:
Chain rule is differentiating what I call "embedded functions".

Composite functions is their real name, btw.
 
  • #8
*mental note to self* TERMINOLOGY!
I knew it, but did not recall on the spot, thanks everyone!
 

1. What are partial fractions?

Partial fractions are a method used in mathematics to decompose a rational function into simpler fractions. This allows for easier integration and manipulation of the function.

2. How do you find the partial fraction decomposition of a rational function?

To find the partial fraction decomposition of a rational function, you must first factor the denominator of the function into its irreducible factors. Then, you set up a system of equations using the coefficients of each factor and solve for the unknown constants. The resulting fractions will make up the partial fraction decomposition of the original function.

3. What is the chain rule?

The chain rule is a rule in calculus that allows for the differentiation of composite functions. It states that the derivative of a composite function is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

4. How do you use the chain rule to differentiate a composite function?

To use the chain rule, you must first identify the outer function and the inner function. Then, you take the derivative of the outer function and evaluate it at the inner function. Finally, you multiply this result by the derivative of the inner function.

5. Can the chain rule be applied to any type of function?

Yes, the chain rule can be applied to any type of function, as long as it is a composite function. This means that it is made up of one function inside another function. The chain rule is a fundamental concept in calculus and is used in various applications, such as in finding rates of change and optimization.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
816
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
357
  • Introductory Physics Homework Help
Replies
15
Views
243
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top