Derivative of cos^(n-1) x

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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 text book 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]
 

Answers and Replies

  • #2
HallsofIvy
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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 text book 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
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Doh, the chain rule.

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

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