How does the product and chain rule apply to this problem?

In summary, the conversation is about a person seeking help understanding a problem in their textbook. The problem involves finding the derivative of a function and the person is struggling to understand how the author gets from one step to another. After some discussion and clarification, it is revealed that the author used factoring to simplify the problem and the person is grateful for the assistance in understanding the solution.
  • #1
JoshHolloway
222
0
Could someone please help me, I do not understand how the author of my textbook gets from one point to another. Here is the problem worked out, after the problem I will explain which part I don't understand.

[tex]f(x)=x(x-4)^3[/tex]
[tex]f'(x)=x[3(x-4)^2]+(x-4)^3[/tex]
[tex]=(x-4)^2(4x-4)[/tex]

I do not understand how the author gets from here:
[tex]f'(x)=x[3(x-4)^2]+(x-4)^3[/tex]
to here:
[tex]=(x-4)^2(4x-4)[/tex]
 
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  • #2
When I do it it looks like this:
[tex]f'(x)=x[3(x-4)^2]+(x-4)^3[/tex]
[tex]=x[3(x^2-8x+16)]+(x-4)(x^2-8x+16)[/tex]
[tex]=x(3x^2-24x+48)+(x^3-8x^2+16x-4x^2+32x-64)[/tex]
[tex]=3x^3-24x^2+48x+x^3-8x^2+16x-4x^2+32x-64[/tex]
[tex]=4x^3-36x^2+96x-64[/tex]
[tex]=4(x^3-9x^2+24x-16)[/tex]
 
  • #3
I am obviously doing something wrong, but I can't seem to figure it out. It will help me a great deal when I figure out what I am doing wrong in these types of problems, so any help is greatly apreciated.
 
  • #4
I believe it was done by factoring [tex](x-4)^2[/tex] out of the second line.

So, you end up with: [tex](x-4)^2[3x+(x-4)][/tex] which reduces to the final expression.
 
  • #5
I don not understand how you can factor [tex](x-4)^2[/tex] out of the second line.
 
  • #6
The square brackets in the second line are redundant and may be part of what is confusing you. That is:

[tex]x[3(x-4)^2] = 3x(x-4)^2[/tex]

Now, rewrite the second line and see if it makes more sense.
 
  • #7
Jimminy Christmas!
It's that easy. Man, I can't believe that. I always mess stupid things up like that. Now it is time for me to find the second derivative.
 
  • #8
I'm glad you understand it. :smile: That sort of thing used to confuse me all the time, and it still does on occasion. The only way to get better at "seeing" those quick solutions is with lots of practice.
 
  • #9
I am learning that the hard way jma. By the way, thanks for the help, I would have never seen that without the assistance.
 

1. How does the product rule work and when is it used?

The product rule is a mathematical formula used to find the derivative of a product of two functions. It is used when the function is a product of two or more simpler functions that cannot be differentiated using basic rules.

2. What is the chain rule and why is it important?

The chain rule is a method for finding the derivative of a composite function, where one function is applied to the output of another function. It is important because it allows us to find the derivative of complex functions by breaking them down into simpler functions.

3. Can the product rule and chain rule be used together?

Yes, the product rule and chain rule can be used together when the function is a product of two or more composite functions. In this case, the chain rule is applied to each individual function and then the product rule is used to find the final derivative.

4. Are there any common mistakes made when applying the product and chain rule?

Yes, some common mistakes include forgetting to apply the chain rule, mixing up the order of operations, and not simplifying the final derivative. It is important to carefully follow the steps of each rule and simplify the final answer.

5. Why do we use the product and chain rule instead of other differentiation methods?

The product and chain rule are used when the function cannot be differentiated using basic rules, such as the power rule or quotient rule. These rules allow us to find the derivative of more complex functions and are essential in many areas of mathematics and science.

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