# Math Help - Strange differential problem?

1. ## Strange 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?

2. ## Re: Strange differential problem?

There's no variable $x$ in the given function, so that means ...

modify

4. ## Re: 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: $dy=f'(x)\cdot dx$)

5. ## Re: Strange differential problem?

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

I am sorry...

6. ## Re: Strange differential problem?

Do you know the chain rule? But write:
$y=\frac{1}{3}\cdot \left[\frac{\cos(6\pi (x-1))}{2}\right]=\frac{1}{6}\cdot \left[\cos(6\pi (x-1))\right]$

But it's still not clear, you have to be more clear with: $\cos(6\pi(x-1))$ or $\cos(6\pi\cdot x-1)$?

7. ## Re: 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.

8. ## Re: Strange differential problem?

Either way you'll have to use the chain rule for this one.

9. ## Re: Strange differential problem?

Originally Posted by homeylova223
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.
is this it?

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

you really need to learn Latex ...

Modified

11. ## Re: Strange differential problem?

Yes skeeter is correct.

I have to apply the chain rule.

Can anyone show me the first step?

12. ## Re: Strange differential problem?

If you want to use the chain rule here:
$D\left[\cos\left(\frac{6\pi x-1}{2}\right)\right]=-\sin\left(\frac{6\pi x-1}{2}\right)\cdot D\left(\frac{6\pi x-1}{2}\right)$. Can you go further? You don't have to use the quotienrule, just put $\frac{1}{2}$ outside, because it's a constant.

13. ## Re: Strange differential problem?

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

would this be correct?

14. ## Re: Strange differential problem?

That's not correct, it has to be:
$D\left(\frac{6\pi x-1}{2}\right)=\frac{1}{2}\cdot D\left(6\pi x-1\right)=\frac{1}{2}\cdot 6\pi=3\pi$

Can you now finish the differential of the given function.

15. ## Re: Strange differential problem?

$y = \frac{1}{3} \cos\left(3\pi x - \frac{1}{2}\right)$

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

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