Strange differential problem?

**homeylova223** · August 4th 2011, 01:52 PM

Find the differential dy of the given function

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

Can anyone help me with this problem?

**Siron** · August 4th 2011, 01:56 PM

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

**homeylova223** · August 4th 2011, 01:57 PM

modify

**Siron** · August 4th 2011, 02:00 PM

That means you're taking the differential of a constant number, and what's the differential of a constant number?
(in general: $dy=f'(x)\cdot dx$ )

**homeylova223** · August 4th 2011, 02:10 PM

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...

**Siron** · August 4th 2011, 02:16 PM

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 $\cos(6\pi\cdot x-1)$ ?

**homeylova223** · August 4th 2011, 02:27 PM

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.

**pickslides** · August 4th 2011, 02:32 PM

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

**skeeter** · August 4th 2011, 02:32 PM

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 ...

**Siron** · August 4th 2011, 02:36 PM

Modified

**homeylova223** · August 4th 2011, 02:36 PM

Yes skeeter is correct.

I have to apply the chain rule.

Can anyone show me the first step?

**Siron** · August 4th 2011, 02:38 PM

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.

**homeylova223** · August 4th 2011, 02:44 PM

Alright following Siron adivce

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

would this be correct?

**Siron** · August 4th 2011, 02:59 PM

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.

**skeeter** · August 4th 2011, 03:06 PM

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

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

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

Thread: Strange differential problem?

LinkBack

Thread Tools

Display

Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Re: Strange differential problem?

Similar Math Help Forum Discussions

A strange integration problem

Strange Problem has me stumped

strange log max&min problem

Strange Problem

Strange Integration Problem

Search Tags