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

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- Dec 5th 2012, 07:43 AMameerulislamdifference 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 - Dec 5th 2012, 07:56 AMtopsquarkRe: difference between cos^3 x and cos x^3?
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 - Dec 5th 2012, 10:42 AMskeeterRe: 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 $\displaystyle \cos(x \cdot x \cdot x)$.

- Dec 6th 2012, 06:32 AMameerulislamRe: 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. - Dec 6th 2012, 06:46 AMPlatoRe: difference between cos^3 x and cos x^3?
- Dec 6th 2012, 06:50 AMameerulislamRe: difference between cos^3 x and cos x^3?
- Dec 6th 2012, 08:16 AMHallsofIvyRe: difference between cos^3 x and cos x^3?
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. - Dec 6th 2012, 09:56 AMameerulislamRe: difference between cos^3 x and cos x^3?
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 :? - Dec 6th 2012, 10:21 AMHallsofIvyRe: 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": $\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}$. - Dec 6th 2012, 01:41 PMskeeterRe: difference between cos^3 x and cos x^3?