# Homework Help: Derivative of a Sine Function Problem Confusion

1. Oct 12, 2009

### Loppyfoot

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

If f(x) = sin²(3-x), then f ' (0) = ____ .

A. -2cos3
B. -2sin3cos3
C. 6cos3
D. 2sin3cos3
E. 6sin3cos3

2. Relevant equations

Derivative of Sinx= cosx.

3. The attempt at a solution

I cannot seem to figure out what the derivative would be with this function.

sin(3-x) * sin(3-x) = Cos(3-x) * Cos(3-x), then plug in 0 for x?

Is there a simpler way using "u" substitution?

Thanks!

2. Oct 12, 2009

### Staff: Mentor

What are you doing in the next line? Does it have anything to do with calculus? I don't see any indication that it does.
Do you know about the chain rule? That's what you need here. Your function could be written this way f(x) = [sin(3 - x)]2.

3. Oct 12, 2009

### Loppyfoot

so, would [ sin(3-x)²] = 2[sin(3-x)] * cos(3-x) *-1?

I think what is confusing me is the square of the function.

4. Oct 12, 2009

### Staff: Mentor

Almost. You're omitting the fact that your taking a derivative.
d/dx[ sin(3-x)²] = 2sin(3-x) * cos(3-x) *(-1) = -2sin(3 - x) cos(3 - x)

5. Oct 12, 2009

### Loppyfoot

Right. Thank you Sir. I know on the AP Calculus exam, the graders take off for not writing limit notation, derivative notation, etc. Thank you for your help, and I definitely understand this chain rule of trigononemtric functions much better.

6. Oct 12, 2009

### Staff: Mentor

You're welcome. Glad to be of help.

7. Oct 12, 2009

### HallsofIvy

Yes, that is correct. That is using the "chain rule".

And note that your earlier "sin(3-x) * sin(3-x) = Cos(3-x) * Cos(3-x)" was wrong for a variety of reasons:
1) You really mean (sin(3-x)*sin(3-x))' on the left. Don't write "=" for things that are not equal!

2) You haven't multiplied by the derivative of 3- x as you should.

3) Most importantly, the derivative of f(x)*f(x) is NOT " f'(x)*f'(x)". It is, by the product rule, f'(x)*f(x)+ f(x)*f'(x) which is equal to 2f(x)f'(x), exactly what applying the chain rule to f'(x)^2 would give you.