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

**ameerulislam** · December 5th 2012, 07:43 AM

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

**topsquark** · December 5th 2012, 07:56 AM

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 $cos^3(x) = cos(x) \cdot cos(x) \cdot cos(x)$ . 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

**skeeter** · December 5th 2012, 10:42 AM

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

**ameerulislam** · December 6th 2012, 06:32 AM

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

Thanks again.

**Plato** · December 6th 2012, 06:46 AM

Originally Posted by ameerulislam

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 $\cos(\sqrt{5x})\text{ or }\sqrt{\cos(5x)}\text{ or even }\cos^5(\sqrt{x})$ .

**ameerulislam** · December 6th 2012, 06:50 AM

Originally Posted by Plato

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 $\sqrt{\cos(5x)}$

**HallsofIvy** · December 6th 2012, 08:16 AM

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 $cos^5(x^{1/2})$ , 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.

**ameerulislam** · December 6th 2012, 09:56 AM

Originally Posted by HallsofIvy

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 :?

**HallsofIvy** · December 6th 2012, 10:21 AM

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": $\frac{dy}{dx}= \frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}$ .

**skeeter** · December 6th 2012, 01:41 PM

Originally Posted by ameerulislam

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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