Finding the time derivative of a trigonometric function

[An1MuS]

Homework Statement

Finding the time derivative of [itex]sin^2( \alpha )[/itex], knowing that [itex] \dot \alpha ≠ 0[/itex]

Homework Equations

i know that [itex] \frac {d}{dt} sin ( \alpha ) = \dot \alpha cos ( \alpha) [/itex]

The Attempt at a Solution

That should give
[itex] \dot \alpha ^2 cos^2( \alpha ) [/itex]

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

 

[The1337gamer]

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

 

[Nugatory]

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

 

[An1MuS]

Another relevant equation [itex] \frac {d}{dx}x^2 = 2dx [/itex]

[itex] \frac {d} {dt} sin^2 \alpha = 2 \dot \alpha cos \alpha [/itex]

Is this it?

 

[LawrenceC]

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

 

[Nugatory]

Another relevant equation [itex] \frac {d}{dx}x^2 = 2dx [/itex]

[itex] \frac {d} {dt} sin^2 \alpha = 2 \dot \alpha cos \alpha [/itex]

Is this it?

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

[itex] \frac {d} {dx} u^2 = 2u \frac {du} {dx}[/itex]

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