Finding the time derivative of a trigonometric function

1. Jan 6, 2012

An1MuS

1. The problem statement, all variables and given/known data

Finding the time derivative of $sin^2( \alpha )$, knowing that $\dot \alpha ≠ 0$

2. Relevant equations

i know that $\frac {d}{dt} sin ( \alpha ) = \dot \alpha cos ( \alpha)$

3. The attempt at a solution

That should give
$\dot \alpha ^2 cos^2( \alpha )$

But it's not and i'm not sure why.

2. Jan 6, 2012

The1337gamer

The best way to evaluate the derivate would be to use the chain rule. Are you familar with the chain rule?

3. Jan 6, 2012

Staff: Mentor

You have one of the relevant equations; another relevant equation is the equation for the derivative of something squared.

4. Jan 7, 2012

An1MuS

Another relevant equation $\frac {d}{dx}x^2 = 2dx$

$\frac {d} {dt} sin^2 \alpha = 2 \dot \alpha cos \alpha$

Is this it?

5. Jan 7, 2012

LawrenceC

d(x^2)/dx = 2x, not 2dx

6. Jan 7, 2012

Staff: Mentor

as LawrenceC has pointed out, you don't have that expression quite right...

But even after you correct it, that's a special case (u=x) of the more general expression

$\frac {d} {dx} u^2 = 2u \frac {du} {dx}$

which you'll probably find more useful. Pay some attention to 1337's comment about the chain rule too.