1. ## Deriv

Hiya

I need to deriv y=0.25*sin(4pi*T)

Am new to this kinda derving but i think it should be somthing like

0.25*cos(8pt*T)

but i think am way off can anyone please give me a hand.
I also need to find the curving amp. curving period

2. Originally Posted by JBswe
Hiya

I need to deriv y=0.25*sin(4pi*T)

Am new to this kinda derving but i think it should be somthing like

0.25*cos(8pt*T)

but i think am way off can anyone please give me a hand.
I also need to find the curving amp. curving period
This is just the chain rule:

$\displaystyle y=0.25*sin(4\pi*T)$

constant stays out front, sin -> cos then find the derivative of what is inside of sin
$\displaystyle \frac{dy}{dT}=0.25*cos(4\pi T)(\frac{dy}{dt}4\pi T)$

$\displaystyle \frac{dy}{dT}=0.25cos(4\pi T)(4\pi)$

And simplify
$\displaystyle \frac{dy}{dT}=\pi cos(4\pi T)$

Just look at it like this, c is the constant coefficient. f(x) is sin(x) and g(x) is $\displaystyle 4\pi T$ then you are taking the derivative of $\displaystyle c f[g(x)]$ which is $\displaystyle c * f\prime [g(x)] * g\prime (x)$

If any of that was confusing, I can explain it in more detail.

3. Originally Posted by JBswe
Hiya

I need to deriv y=0.25*sin(4pi*T)

Am new to this kinda derving but i think it should be somthing like

0.25*cos(8pt*T)

but i think am way off can anyone please give me a hand.
I also need to find the curving amp. curving period
If $\displaystyle y = a \sin (kT)$ then $\displaystyle \frac{dy}{dT} = k a \cos (kT)$.

In your question a = 0.25 and $\displaystyle k = 4 \pi$.