# Thread: difference between cos^3 x and cos x^3?

1. ## 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?

Thanks

2. ## Re: difference between cos^3 x and cos x^3?

Originally Posted by ameerulislam
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 $\displaystyle cos^3(x) = cos(x) \cdot cos(x) \cdot cos(x)$. However it could also be taken to be $\displaystyle cos(cos(cos(x)))$.

Without parenthesis cos x^3 is $\displaystyle cos(x^3)$.

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

-Dan

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

note that some calculators have the syntax $\displaystyle \cos(x)^3$ to mean $\displaystyle (\cos{x})^3$ , and $\displaystyle \cos(x^3)$ to mean $\displaystyle \cos(x \cdot x \cdot x)$.

4. ## 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.

5. ## Re: difference between cos^3 x and cos x^3?

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

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

But someone else may read it as $\displaystyle \cos(\sqrt{5x})\text{ or }\sqrt{\cos(5x)}\text{ or even }\cos^5(\sqrt{x})$.

6. ## Re: difference between cos^3 x and cos x^3?

Originally Posted by Plato

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

But someone else may read it as $\displaystyle \cos(\sqrt{5x})\text{ or }\sqrt{\cos(5x)}\text{ or even }\cos^5(\sqrt{x})$.
sorry, its $\displaystyle \sqrt{\cos(5x)}$

7. ## Re: difference between cos^3 x and cos x^3?

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

Thanks again.
No! neither $\displaystyle cos^5(x^{1/2})$, which is what I think you mean, nor $\displaystyle cos(5x^{1/2})$, which is what you actually wrote, is equal to $\displaystyle 5 cos(x^{1/2}$. 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, $\displaystyle x^{1/2}= \sqrt{x}= \sqrt{4}= 2$ so that $\displaystyle cos(x^{1/2})= cos(2)= -0.4161$ (approximately) so that $\displaystyle cos^5(x^{1/2})= (-0.4161)^5= -0.01248$, approximately. Of course, $\displaystyle 5cos(x^{1/2})= 5 cos(2)= 5(-.4161)= -2.0807$. Of course, $\displaystyle 5x^{1/2}= 5(2)= 10$ so $\displaystyle cos(5x^{1/2})= cos(10)= -0.8391$.

Those are three different calculations giving three different answers.

8. ## Re: difference between cos^3 x and cos x^3?

Originally Posted by HallsofIvy
No! neither $\displaystyle cos^5(x^{1/2})$, which is what I think you mean, nor $\displaystyle cos(5x^{1/2})$, which is what you actually wrote, is equal to $\displaystyle 5 cos(x^{1/2}$. 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, $\displaystyle x^{1/2}= \sqrt{x}= \sqrt{4}= 2$ so that $\displaystyle cos(x^{1/2})= cos(2)= -0.4161$ (approximately) so that $\displaystyle cos^5(x^{1/2})= (-0.4161)^5= -0.01248$, approximately. Of course, $\displaystyle 5cos(x^{1/2})= 5 cos(2)= 5(-.4161)= -2.0807$. Of course, $\displaystyle 5x^{1/2}= 5(2)= 10$ so $\displaystyle cos(5x^{1/2})= cos(10)= -0.8391$.

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

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

9. ## Re: difference between cos^3 x and cos x^3?

The chain rule! $\displaystyle y= \sqrt{cos(5x)}=(cos(5x))^{1/2}$ which we can write as $\displaystyle u^{1/2}$, with $\displaystyle u= cos(v)$ and $\displaystyle v= 5x$. $\displaystyle \frac{dy}{u}= (1/2)u^{-1/2}$ and $\displaystyle \frac{du}{dv}= -sin(v)$ so that $\displaystyle \frac{dy}{dv}= \frac{dy}{du}\frac{du}{dv}=$$\displaystyle (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 $\displaystyle \frac{dy}{dx}$ apply the chain rule again: $\displaystyle \frac{dy}{dx}= \frac{dy}{dv}\frac{dv}{dx}$. Since v= 5x, $\displaystyle \frac{dv}{dx}= 5$ so $\displaystyle \frac{dy}{dx}= \frac{-sin(v)}{2\sqrt{sin(v)}}(5)= -\frac{5sin(v)}{2\sqrt{sin(v)}}= -\frac{5sin(5x)}{2\sqrt{cos(5x)}}$.

Of course, you could also do that as one "chain rule": $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}$.

10. ## Re: difference between cos^3 x and cos x^3?

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

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

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