Find the differential dy of the given function

y= (1/3) cos((6pi times x-1))/((2))

Can anyone help me with this problem?

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- Aug 4th 2011, 01:52 PMhomeylova223Strange differential problem?
Find the differential dy of the given function

y= (1/3) cos((6pi times x-1))/((2))

Can anyone help me with this problem? - Aug 4th 2011, 01:56 PMSironRe: Strange differential problem?
There's no variable $\displaystyle x$ in the given function, so that means ...

- Aug 4th 2011, 01:57 PMhomeylova223Re: Strange differential problem?
modify

- Aug 4th 2011, 02:00 PMSironRe: Strange differential problem?
That means you're taking the differential of a constant number, and what's the differential of a constant number?

(in general: $\displaystyle dy=f'(x)\cdot dx$) - Aug 4th 2011, 02:10 PMhomeylova223Re: Strange differential problem?
Actually I have to admit I made a typo

y= (1/3) cos((6pi x-1))/((2)) is how it is correctly written

I am sorry... - Aug 4th 2011, 02:16 PMSironRe: Strange differential problem?
Do you know the chain rule? But write:

$\displaystyle y=\frac{1}{3}\cdot \left[\frac{\cos(6\pi (x-1))}{2}\right]=\frac{1}{6}\cdot \left[\cos(6\pi (x-1))\right]$

But it's still not clear, you have to be more clear with: $\displaystyle \cos(6\pi(x-1))$ or $\displaystyle \cos(6\pi\cdot x-1)$? - Aug 4th 2011, 02:27 PMhomeylova223Re: Strange differential problem?
No in my textbook the problem is written

((1/3)) cos ((6pi times x-1))/((2))

The cos is outside the parenthesis

It can be tough being clear sometimes. - Aug 4th 2011, 02:32 PMpickslidesRe: Strange differential problem?
Either way you'll have to use the chain rule for this one.

- Aug 4th 2011, 02:32 PMskeeterRe: Strange differential problem?
- Aug 4th 2011, 02:36 PMhomeylova223Re: Strange differential problem?
Yes skeeter is correct.

I have to apply the chain rule.

Can anyone show me the first step? - Aug 4th 2011, 02:36 PMSironRe: Strange differential problem?
Modified

- Aug 4th 2011, 02:38 PMSironRe: Strange differential problem?
If you want to use the chain rule here:

$\displaystyle D\left[\cos\left(\frac{6\pi x-1}{2}\right)\right]=-\sin\left(\frac{6\pi x-1}{2}\right)\cdot D\left(\frac{6\pi x-1}{2}\right)$. Can you go further? You don't have to use the quotienrule, just put $\displaystyle \frac{1}{2}$ outside, because it's a constant. - Aug 4th 2011, 02:44 PMhomeylova223Re: Strange differential problem?
Alright following Siron adivce

-sin((6pi times x-1))/((2)) times cos(6pi)(-1/2)

would this be correct? - Aug 4th 2011, 02:59 PMSironRe: Strange differential problem?
That's not correct, it has to be:

$\displaystyle D\left(\frac{6\pi x-1}{2}\right)=\frac{1}{2}\cdot D\left(6\pi x-1\right)=\frac{1}{2}\cdot 6\pi=3\pi$

Can you now finish the differential of the given function. - Aug 4th 2011, 03:06 PMskeeterRe: Strange differential problem?
$\displaystyle y = \frac{1}{3} \cos\left(3\pi x - \frac{1}{2}\right)$

$\displaystyle \frac{dy}{dx} = -\frac{1}{3} \sin\left(3\pi x - \frac{1}{2}\right) \cdot 3\pi$

$\displaystyle \frac{dy}{dx} = -\pi \sin\left(3\pi x - \frac{1}{2}\right)$