Can You Derive the Formula for cos^(n-1)(x) Using a Recursive Method?

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In summary, the problem is that the student is trying to differentiate a cosine function but they are not familiar with the differentiation techniques. They are looking for a good example and they found it in Exhibit A. This website provides demonstrations on how to find the formula for the cosine function. The student is rusty on their differentiation techniques and is concerned with their communication marks.
  • #1
Patjamet
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I'm working on a Maths assignment and one of the problem solving questions is to prove:
[tex]\int cos^{8}(x)dx=\frac{1}{8}cos^{7}(x).sin(x)+\frac{7}{48}cos^{5}(x).sin(x)+\frac{35}{192}cos^{3}(x).sin(x)+\frac{35}{128}cos(x).sin(x)+\frac{35}{128}(x)+c[/tex]

I have looked through my textbook and found a good example which will help me, the book uses a thing they are calling a recursive formula. I've done some research and I've seen the same formula from lots of different information sources.

Exhibit A: http://calc101.com/deriving_reduction_2.html

[STRIKE]They have all given demonstrations on how to get the formula, but when I try to do it myself I run into a problem when differentiating the first function [tex]cos^{n-1}(x)[/tex][/STRIKE]

Nevermind I was integrating haha, but still; obviously I am rusty on my differentiation techniques so if someone could check this for me that would be fantastic (mainly concerned with my procedure for communication marks etc.) :)

[tex]y=cos^{n-1}(x)[/tex]Allow cos(x) to equal u[tex]\frac{dy}{dx}=u^{n-1}....1[/tex]


[tex]\frac{du}{dx}=-sin(x)[/tex]

[tex]\frac{du}{-sin(x)}=dx.....2[/tex]Sub 2 into 1[tex]\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}[/tex]

[tex]\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)[/tex]

[tex]\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)[/tex]

But u = cosx

[tex]\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)[/tex]
 
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  • #2
Patjamet said:
I'm working on a Maths assignment and one of the problem solving questions is to prove:
[tex]\int cos^{8}(x)dx=\frac{1}{8}cos^{7}(x).sin(x)+\frac{7}{48}cos^{5}(x).sin(x)+\frac{35}{192}cos^{3}(x).sin(x)+\frac{35}{128}cos(x).sin(x)+\frac{35}{128}(x)+c[/tex]

I have looked through my textbook and found a good example which will help me, the book uses a thing they are calling a recursive formula. I've done some research and I've seen the same formula from lots of different information sources.

Exhibit A: http://calc101.com/deriving_reduction_2.html

[STRIKE]They have all given demonstrations on how to get the formula, but when I try to do it myself I run into a problem when differentiating the first function [tex]cos^{n-1}(x)[/tex][/STRIKE]

Nevermind I was integrating haha, but still; obviously I am rusty on my differentiation techniques so if someone could check this for me that would be fantastic (mainly concerned with my procedure for communication marks etc.) :)

[tex]y=cos^{n-1}(x)[/tex]


Allow cos(x) to equal u


[tex]\frac{dy}{dx}=u^{n-1}....1[/tex]
No, you haven't done the derivative yet. [itex]y= u^{n-1}[/itex]
and then [itex]dy/dx= (n-1)u^{n-2}

[/quote][tex]\frac{du}{dx}=-sin(x)[/tex]

[tex]\frac{du}{-sin(x)}=dx.....2[/tex]
Why are you doing this? Use the chain rule: (du/dx)= (du/dy)(dy/dx)
That gives you [itex])n-1)u^{n-2}(-sin(x))= -(n-2)sin(x)cos^{n-2}(x)


Sub 2 into 1


[tex]\frac{dy}{\frac{du}{-sin(x)}}=u^{n-1}[/tex]

[tex]\frac{dy}{du}=(n-1)u^{n-1-1}.-sin(x)[/tex]

[tex]\frac{dy}{du}=(n-1)u^{n-2}.-sin(x)[/tex]

But u = cosx

[tex]\frac{dy}{du}=(n-1)cos^{n-2}(x).-sin(x)[/tex]

But I see no reason to ask about the "n-1" power. For your problem you want to look at [itex]cos^n(x)[/itex]. Again, letting u= cos(x), [itex]y= u^n[/itex] so [itex]dy/du= n u^{n-1}[/itex] and du/dx= - sin(x). So [itex]dy/dx= (dy/du)(du/dx)= n u^{n-1}(-sin(x))= -n sin(x)cos^{n-1}(x)[/itex].

With a little practice, you should be able to do that with actually writing down the "u" substitution: to differentiate [itex]cos^n(x)[/itex] think "The "outer function" is a power so the derivative is [itex]n cos^{n-1}(x)[/itex] and then I multiply by its derivative, -sin(x)".
 
  • #3
Doh, the chain rule.

Thank you very much, looks like I have a lot of revision to do!
 

1. What is the general formula for finding the derivative of cos^(n-1) x?

The general formula for finding the derivative of cos^(n-1) x is -sin(n-1)x * (n-1) * cos^(n-2) x * sin x.

2. Can the derivative of cos^(n-1) x be simplified further?

Yes, the derivative can be written as (n-1) * cos^(n-2) x * cos x.

3. How do you use the chain rule to find the derivative of cos^(n-1) x?

To use the chain rule, we first rewrite cos^(n-1) x as (cos x)^n-1. Then, we apply the power rule and the chain rule to get (n-1) * cos^(n-2) x * -sin x * cos x.

4. What is the derivative of cos^2 x?

The derivative of cos^2 x is -2 * cos x * sin x.

5. Are there any special cases to consider when finding the derivative of cos^(n-1) x?

Yes, when n = 1, the derivative of cos^(n-1) x is simply -sin x. Additionally, when n = 0, the derivative is undefined.

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