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Math Help - difference between cos^3 x and cos x^3?

  1. #1
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    difference between cos^3 x and cos x^3?

    So, what is the difference between cos^3 x, cos x^3 and (cos x)^3 or is there even a difference in between them?

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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by ameerulislam View Post
    So, what is the difference between cos^3 x, cos x^3 and (cos x)^3 or is there even a difference in between them?

    Thanks
    Typically cos^3 x is cos^3(x) = cos(x) \cdot cos(x) \cdot cos(x). However it could also be taken to be cos(cos(cos(x))).

    Without parenthesis cos x^3 is cos(x^3).

    And (cos x)^3 is simply cos^3(x) = cos(x) \cdot cos(x) \cdot cos(x).

    -Dan
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    Re: difference between cos^3 x and cos x^3?

    note that some calculators have the syntax \cos(x)^3 to mean (\cos{x})^3 , and \cos(x^3) to mean \cos(x \cdot x \cdot x).
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    Re: difference between cos^3 x and cos x^3?

    ok what about cos5x^1/2 , can I move 5 like this 5 cos x^1/2 ?


    Thanks again.
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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by ameerulislam View Post
    ok what about cos5x^1/2 , can I move 5 like this 5 cos x^1/2 ?

    Please, please learn to use grouping symbols.

    I just have to guess that you mean cos(5x^1/2) or \cos(5\sqrt{x})~?

    But someone else may read it as \cos(\sqrt{5x})\text{ or }\sqrt{\cos(5x)}\text{ or even }\cos^5(\sqrt{x}).
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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by Plato View Post
    Please, please learn to use grouping symbols.

    I just have to guess that you mean cos(5x^1/2) or \cos(5\sqrt{x})~?

    But someone else may read it as \cos(\sqrt{5x})\text{ or }\sqrt{\cos(5x)}\text{ or even }\cos^5(\sqrt{x}).
    sorry, its \sqrt{\cos(5x)}
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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by ameerulislam View Post
    ok what about cos5x^1/2 , can I move 5 like this 5 cos x^1/2 ?


    Thanks again.
    No! neither cos^5(x^{1/2}), which is what I think you mean, nor cos(5x^{1/2}), which is what you actually wrote, is equal to 5 cos(x^{1/2}. The first means "find the square root of x, then take the cosine of that number, then multiply it by itself 5 times". The second means "find the square root of x, multiply that by 5, then take the cosine". The last means "find the square root of x, find the cosine of it, then multiply by 5".

    For example, if x= 4, x^{1/2}= \sqrt{x}= \sqrt{4}= 2 so that cos(x^{1/2})= cos(2)= -0.4161 (approximately) so that cos^5(x^{1/2})= (-0.4161)^5= -0.01248, approximately. Of course, 5cos(x^{1/2})= 5 cos(2)= 5(-.4161)= -2.0807. Of course, 5x^{1/2}= 5(2)= 10 so cos(5x^{1/2})= cos(10)= -0.8391.

    Those are three different calculations giving three different answers.
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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by HallsofIvy View Post
    No! neither cos^5(x^{1/2}), which is what I think you mean, nor cos(5x^{1/2}), which is what you actually wrote, is equal to 5 cos(x^{1/2}. The first means "find the square root of x, then take the cosine of that number, then multiply it by itself 5 times". The second means "find the square root of x, multiply that by 5, then take the cosine". The last means "find the square root of x, find the cosine of it, then multiply by 5".

    For example, if x= 4, x^{1/2}= \sqrt{x}= \sqrt{4}= 2 so that cos(x^{1/2})= cos(2)= -0.4161 (approximately) so that cos^5(x^{1/2})= (-0.4161)^5= -0.01248, approximately. Of course, 5cos(x^{1/2})= 5 cos(2)= 5(-.4161)= -2.0807. Of course, 5x^{1/2}= 5(2)= 10 so cos(5x^{1/2})= cos(10)= -0.8391.

    Those are three different calculations giving three different answers.
    ok here where I'm really troubled.. it's a derivative problem

    {d/dx}\sqrt {cos(5x)}
    The result I'm getting is - {sin(5x)/2\sqrt{cos(5x)}}
    while the actual result is
    - {5 sin(5x)/2\sqrt{cos(5x)}}
    I can't figure out where this extra 5 in front of sin came from :?
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    Re: difference between cos^3 x and cos x^3?

    The chain rule! y= \sqrt{cos(5x)}=(cos(5x))^{1/2} which we can write as u^{1/2}, with u= cos(v) and v= 5x. \frac{dy}{u}= (1/2)u^{-1/2} and \frac{du}{dv}= -sin(v) so that \frac{dy}{dv}= \frac{dy}{du}\frac{du}{dv}=  (1/2)u^{-1/2}(-sin(v))= \frac{-sin(v)}{2\sqrt{cos(v)}} which is very similar to what you have except that it is still in terms of v, not u. To get \frac{dy}{dx} apply the chain rule again: \frac{dy}{dx}= \frac{dy}{dv}\frac{dv}{dx}. Since v= 5x, \frac{dv}{dx}= 5 so \frac{dy}{dx}= \frac{-sin(v)}{2\sqrt{sin(v)}}(5)= -\frac{5sin(v)}{2\sqrt{sin(v)}}= -\frac{5sin(5x)}{2\sqrt{cos(5x)}}.

    Of course, you could also do that as one "chain rule": \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}.
    Last edited by HallsofIvy; December 6th 2012 at 10:24 AM.
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    Re: difference between cos^3 x and cos x^3?

    Quote Originally Posted by ameerulislam View Post
    ok here where I'm really troubled.. it's a derivative problem

    {d/dx}\sqrt {cos(5x)}
    ... another case of the need to post the entire problem from the start.
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