y(t)=(3/2)*sin(12*pi*t)

y't=(3/2)*cos(12*pi*t)

Is that correct?

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- Sep 18th 2008, 10:05 AMNo Logic SenseDifferential help!
y(t)=(3/2)*sin(12*pi*t)

y't=(3/2)*cos(12*pi*t)

Is that correct? - Sep 18th 2008, 10:12 AMChris L T521
- Sep 18th 2008, 10:14 AMMoo
- Sep 18th 2008, 10:17 AMNo Logic Sense
Okay, let me try again:

y't=(3/2)*cos(t)

Is that correct? - Sep 18th 2008, 10:24 AMNo Logic Sense
No wait, I can do it!

If 12 * pi * t = u

and sin(u) = y(t)

du/dy = -sin(u)

dt/dy = t

y'(t)=(3/2) * t * -sin(12*pi*t)

Is that now correct? :) - Sep 18th 2008, 10:29 AMMoo
Hmmm... There is a confusion... Pardon me if it is my fault :(

Get back to my formula.

$\displaystyle g(t)=12 \pi t \implies g'(t)=12 \pi$

$\displaystyle f(x)=\sin(x) \implies f'(x)=\cos(x)$

So the formula give :

$\displaystyle y'(t)=\frac 32 \cdot [f(g(t))]'=\frac 32 \cdot g'(t) \cdot f'(g(t))=\frac 32 \cdot 12 \pi \cdot \cos(12 \pi t)=\dots$ - Sep 18th 2008, 10:31 AMNo Logic Sense
Oh yeah, you're right.

No don't worry it wasn't your fault. It has just been ages since I did the whole diffential thing so I thought 12 * pi would dissapear, but of course it's t.

Thanks for the answers. :)