1. ## Differential help!

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

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

Is that correct?

2. Originally Posted by No Logic Sense
y(t)=(3/2)*sin(12*pi*t)

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

Is that correct?
Close... You forgot to apply the chain rule! You need to multiply $\frac{d}{\,dt}\left[12\pi t\right]=\dots$ to your answer.

--Chris

3. Hello,
Originally Posted by No Logic Sense
y(t)=(3/2)*sin(12*pi*t)

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

Is that correct?
Nope. You forgot to use the chain rule

$[f(g(x))]'=g'(x)f'(g(x))$

Here, $g(t)=12 \pi t$ and $f(x)=\sin(x)$

Basically, you'll have to multiply by the derivative of 12 pi t, with respect to t

4. Okay, let me try again:

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

Is that correct?

5. 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?

6. Originally Posted by No 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?
Hmmm... There is a confusion... Pardon me if it is my fault

Get back to my formula.

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

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

So the formula give :

$y'(t)=\frac 32 \cdot [f(g(t))]'=\frac 32 \cdot g'(t) \cdot f'(g(t))=\frac 32 \cdot 12 \pi \cdot \cos(12 \pi t)=\dots$

7. 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.