# Thread: Derivative of Sin function Problem

1. ## Derivative of Sin function Problem

If f(x)= sin²(3-x), then dy/dx(0)= ?

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

This is a questions from my AP Calculus BC class, and I can't seem to get the correct answer. Any Thoughts?

Thanks!

2. Originally Posted by r2d2
If f(x)= sin²(3-x), then dy/dx(0)= ?
Any Thoughts?

Thanks!
Sure, start by differentiating $f(x)$.

It may be the case that you are confused because the problem contains two forms of notation.

Understand that $f'(x)=\frac{dy}{dx}$. The fact that this question would mix and match notation is a strong indication that your teachers are crap. But, don't let it get ya down.

So, you are asked to find $f'(0)$...

3. so f'(x)= sin²(3-x)= sin(3-x) * (sin(3-x) = cos(3-x) * cos(3-x)

Then plug in f'(0) to get cos(3) * cos(3)?

Would that be the way to go about this problem? I think I am missing an important trigonometric derivative point.... Any Thoughts?

4. Rewrite as (sin(3-x))²

Then use chain rule:

2(sin(3-x)) (cos(3-x)) (-1)

,

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### differentiation of sin²3x

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