# Homework Help: 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.